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Function that multiplies each element of one list with four elements of second list in a loop to get a new list
Getting the last element of a listHow do I get the number of elements in a list?Splitting a list into N parts of approximately equal lengthWhy do I get an UnsupportedOperationException when trying to remove an element from a List?Call int() function on every list element?List order manipulating in PythonPython: How do I add a suffix to the end of each list element?Python lists intersection/comparison with (two) loopsMultiply all combinations of two listsHow to efficiently find out the maximum value below a threshold?
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0
I have two lists,one of length 40 and the other of length 10.I want to multiply first four elements of the 40 list with the first element of the second list and loop that for the entire first list of 40 to a get a new list which is the product of these two.
Any suggestions on how to go about this?
from itertools import zip_longest
def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
args = [iter(iterable)] * n
return zip_longest(*args, fillvalue=fillvalue)
v=[[a1/ b1 for a1 in a4] for a4, b1 in zip(grouper(active, 4), passive)]
active=[56.93977737426758,
54.12062072753906,
54.89398765563965,
55.214101791381836,
54.29464149475098,
53.80845832824707,
54.46353721618652,
54.49761962890625,
53.01671028137207,
53.872962951660156,
53.156455993652344,
53.20746994018555,
52.529762268066406,
56.03120040893555,
54.122426986694336,
55.83853149414063,
53.51207160949707,
54.82537269592285,
53.569284439086914,
53.5296745300293,
54.354637145996094,
54.313310623168945,
53.26720809936523,
54.64541053771973,
55.00912475585938,
55.093666076660156,
55.138763427734375,
54.19987297058106,
54.07197189331055,
53.18226623535156,
53.656246185302734,
54.97188377380371,
55.28757095336914,
54.08882141113281,
53.08153915405274,
53.61944770812988,
53.15986633300781,
53.53702735900879,
53.32623863220215,
52.01462173461914]
passive= [54.46392059326172,
52.37292861938477,
51.95756149291992,
53.40110778808594,
54.46831512451172,
56.04657173156738,
57.74487495422363,
53.75052452087402,
56.246402740478516,
55.15713691711426]
My current output is a list of 10.I want a list of 40.I want to take divide the first four elements of active with first element of passive...and so on.In the end I want a new list of 40 elements and not 10.
example [active1/passive1,active2/passive1,active3/passive1....active40/passive10]
[[1.0454586587604597,
0.9936967470945319,
1.0078963662125922,
1.0137739110579735],
[1.0366928664488493,
1.0274097658218595,
1.0399177333006342,
1.040568497228071],
[1.020384882546816,
1.0368647296698332,
1.0230744951511204,
1.0240563338877238],
[0.9836830066620091,
1.0492516490722763,
1.013507569945381,
1.045643691807429],
[0.9824440408551517,
1.0065553261670555,
0.9834944282126284,
0.982767218109524],
[0.9698119879004422,
0.9690746274955083,
0.9504097477092123,
0.9750000553011796],
[0.9526234977470646,
0.954087546649548,
0.9548685224696527,
0.9386092361191739],
[1.005980357871888,
0.9894278559960512,
0.9982460015709302,
1.0227227411047006],
[0.9829530113857514,
0.9616405454531767,
0.9437321600631335,
0.9532955903958973],
[0.9637894441999811,
0.9706273811757119,
0.9668057773255447,
0.9430261366318299]]
python list multiplying
asked Mar 25 at 7:53
mashedpotetomashedpoteto
105
|
show 2 more comments
0
I have two lists,one of length 40 and the other of length 10.I want to multiply first four elements of the 40 list with the first element of the second list and loop that for the entire first list of 40 to a get a new list which is the product of these two.
Any suggestions on how to go about this?
from itertools import zip_longest
def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
args = [iter(iterable)] * n
return zip_longest(*args, fillvalue=fillvalue)
v=[[a1/ b1 for a1 in a4] for a4, b1 in zip(grouper(active, 4), passive)]
active=[56.93977737426758,
54.12062072753906,
54.89398765563965,
55.214101791381836,
54.29464149475098,
53.80845832824707,
54.46353721618652,
54.49761962890625,
53.01671028137207,
53.872962951660156,
53.156455993652344,
53.20746994018555,
52.529762268066406,
56.03120040893555,
54.122426986694336,
55.83853149414063,
53.51207160949707,
54.82537269592285,
53.569284439086914,
53.5296745300293,
54.354637145996094,
54.313310623168945,
53.26720809936523,
54.64541053771973,
55.00912475585938,
55.093666076660156,
55.138763427734375,
54.19987297058106,
54.07197189331055,
53.18226623535156,
53.656246185302734,
54.97188377380371,
55.28757095336914,
54.08882141113281,
53.08153915405274,
53.61944770812988,
53.15986633300781,
53.53702735900879,
53.32623863220215,
52.01462173461914]
passive= [54.46392059326172,
52.37292861938477,
51.95756149291992,
53.40110778808594,
54.46831512451172,
56.04657173156738,
57.74487495422363,
53.75052452087402,
56.246402740478516,
55.15713691711426]
My current output is a list of 10.I want a list of 40.I want to take divide the first four elements of active with first element of passive...and so on.In the end I want a new list of 40 elements and not 10.
example [active1/passive1,active2/passive1,active3/passive1....active40/passive10]
[[1.0454586587604597,
0.9936967470945319,
1.0078963662125922,
1.0137739110579735],
[1.0366928664488493,
1.0274097658218595,
1.0399177333006342,
1.040568497228071],
[1.020384882546816,
1.0368647296698332,
1.0230744951511204,
1.0240563338877238],
[0.9836830066620091,
1.0492516490722763,
1.013507569945381,
1.045643691807429],
[0.9824440408551517,
1.0065553261670555,
0.9834944282126284,
0.982767218109524],
[0.9698119879004422,
0.9690746274955083,
0.9504097477092123,
0.9750000553011796],
[0.9526234977470646,
0.954087546649548,
0.9548685224696527,
0.9386092361191739],
[1.005980357871888,
0.9894278559960512,
0.9982460015709302,
1.0227227411047006],
[0.9829530113857514,
0.9616405454531767,
0.9437321600631335,
0.9532955903958973],
[0.9637894441999811,
0.9706273811757119,
0.9668057773255447,
0.9430261366318299]]
python list multiplying
asked Mar 25 at 7:53
mashedpotetomashedpoteto
105
Sample lists with what you have already tried would boost your chances of getting valid answers.
– DirtyBit
Mar 25 at 7:54
Can we see some code? Give it a shot.
– Aran-Fey
Mar 25 at 7:55
Multiply them how? 1*1, 2*1, 3*1, 4*1 or 1*2*3*4*1? (Both lists being natural numbers sequences starting in 1)
– Netwave
Mar 25 at 7:58
a=[1,2,3,.....40] len(a)=40 b=[1,2,....10] len(b)=10
– mashedpoteto
Mar 25 at 8:14
I want to make two lists,one by multiplying & other by dividing ->every first four element of a[ ] with the every element of b [ ] in loop. So something like : [1,2,3,4]/1 -> will be the first element of my new list. The second will be [5,6,7,8]/2 and so on.I hope this makes clear. The last element of the new set would be [37,38,39,40]/10. The above example was for division operator.I want to repeat the same thing with multiplication.
– mashedpoteto
Mar 25 at 8:19
|
show 2 more comments
0
0
0
I have two lists,one of length 40 and the other of length 10.I want to multiply first four elements of the 40 list with the first element of the second list and loop that for the entire first list of 40 to a get a new list which is the product of these two.
Any suggestions on how to go about this?
from itertools import zip_longest
def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
args = [iter(iterable)] * n
return zip_longest(*args, fillvalue=fillvalue)
v=[[a1/ b1 for a1 in a4] for a4, b1 in zip(grouper(active, 4), passive)]
active=[56.93977737426758,
54.12062072753906,
54.89398765563965,
55.214101791381836,
54.29464149475098,
53.80845832824707,
54.46353721618652,
54.49761962890625,
53.01671028137207,
53.872962951660156,
53.156455993652344,
53.20746994018555,
52.529762268066406,
56.03120040893555,
54.122426986694336,
55.83853149414063,
53.51207160949707,
54.82537269592285,
53.569284439086914,
53.5296745300293,
54.354637145996094,
54.313310623168945,
53.26720809936523,
54.64541053771973,
55.00912475585938,
55.093666076660156,
55.138763427734375,
54.19987297058106,
54.07197189331055,
53.18226623535156,
53.656246185302734,
54.97188377380371,
55.28757095336914,
54.08882141113281,
53.08153915405274,
53.61944770812988,
53.15986633300781,
53.53702735900879,
53.32623863220215,
52.01462173461914]
passive= [54.46392059326172,
52.37292861938477,
51.95756149291992,
53.40110778808594,
54.46831512451172,
56.04657173156738,
57.74487495422363,
53.75052452087402,
56.246402740478516,
55.15713691711426]
My current output is a list of 10.I want a list of 40.I want to take divide the first four elements of active with first element of passive...and so on.In the end I want a new list of 40 elements and not 10.
example [active1/passive1,active2/passive1,active3/passive1....active40/passive10]
[[1.0454586587604597,
0.9936967470945319,
1.0078963662125922,
1.0137739110579735],
[1.0366928664488493,
1.0274097658218595,
1.0399177333006342,
1.040568497228071],
[1.020384882546816,
1.0368647296698332,
1.0230744951511204,
1.0240563338877238],
[0.9836830066620091,
1.0492516490722763,
1.013507569945381,
1.045643691807429],
[0.9824440408551517,
1.0065553261670555,
0.9834944282126284,
0.982767218109524],
[0.9698119879004422,
0.9690746274955083,
0.9504097477092123,
0.9750000553011796],
[0.9526234977470646,
0.954087546649548,
0.9548685224696527,
0.9386092361191739],
[1.005980357871888,
0.9894278559960512,
0.9982460015709302,
1.0227227411047006],
[0.9829530113857514,
0.9616405454531767,
0.9437321600631335,
0.9532955903958973],
[0.9637894441999811,
0.9706273811757119,
0.9668057773255447,
0.9430261366318299]]
python list multiplying
asked Mar 25 at 7:53
mashedpotetomashedpoteto
105
I have two lists,one of length 40 and the other of length 10.I want to multiply first four elements of the 40 list with the first element of the second list and loop that for the entire first list of 40 to a get a new list which is the product of these two.
Any suggestions on how to go about this?
from itertools import zip_longest
def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
args = [iter(iterable)] * n
return zip_longest(*args, fillvalue=fillvalue)
v=[[a1/ b1 for a1 in a4] for a4, b1 in zip(grouper(active, 4), passive)]
active=[56.93977737426758,
54.12062072753906,
54.89398765563965,
55.214101791381836,
54.29464149475098,
53.80845832824707,
54.46353721618652,
54.49761962890625,
53.01671028137207,
53.872962951660156,
53.156455993652344,
53.20746994018555,
52.529762268066406,
56.03120040893555,
54.122426986694336,
55.83853149414063,
53.51207160949707,
54.82537269592285,
53.569284439086914,
53.5296745300293,
54.354637145996094,
54.313310623168945,
53.26720809936523,
54.64541053771973,
55.00912475585938,
55.093666076660156,
55.138763427734375,
54.19987297058106,
54.07197189331055,
53.18226623535156,
53.656246185302734,
54.97188377380371,
55.28757095336914,
54.08882141113281,
53.08153915405274,
53.61944770812988,
53.15986633300781,
53.53702735900879,
53.32623863220215,
52.01462173461914]
passive= [54.46392059326172,
52.37292861938477,
51.95756149291992,
53.40110778808594,
54.46831512451172,
56.04657173156738,
57.74487495422363,
53.75052452087402,
56.246402740478516,
55.15713691711426]
My current output is a list of 10.I want a list of 40.I want to take divide the first four elements of active with first element of passive...and so on.In the end I want a new list of 40 elements and not 10.
example [active1/passive1,active2/passive1,active3/passive1....active40/passive10]
[[1.0454586587604597,
0.9936967470945319,
1.0078963662125922,
1.0137739110579735],
[1.0366928664488493,
1.0274097658218595,
1.0399177333006342,
1.040568497228071],
[1.020384882546816,
1.0368647296698332,
1.0230744951511204,
1.0240563338877238],
[0.9836830066620091,
1.0492516490722763,
1.013507569945381,
1.045643691807429],
[0.9824440408551517,
1.0065553261670555,
0.9834944282126284,
0.982767218109524],
[0.9698119879004422,
0.9690746274955083,
0.9504097477092123,
0.9750000553011796],
[0.9526234977470646,
0.954087546649548,
0.9548685224696527,
0.9386092361191739],
[1.005980357871888,
0.9894278559960512,
0.9982460015709302,
1.0227227411047006],
[0.9829530113857514,
0.9616405454531767,
0.9437321600631335,
0.9532955903958973],
[0.9637894441999811,
0.9706273811757119,
0.9668057773255447,
0.9430261366318299]]
python list multiplying
python list multiplying
asked Mar 25 at 7:53
mashedpotetomashedpoteto
105
asked Mar 25 at 7:53
mashedpotetomashedpoteto
105
edited Mar 25 at 8:52
mashedpoteto
asked Mar 25 at 7:53
mashedpotetomashedpoteto
105
asked Mar 25 at 7:53
mashedpotetomashedpoteto
105
asked Mar 25 at 7:53
mashedpotetomashedpoteto
105
105
Sample lists with what you have already tried would boost your chances of getting valid answers.
– DirtyBit
Mar 25 at 7:54
Can we see some code? Give it a shot.
– Aran-Fey
Mar 25 at 7:55
Multiply them how? 1*1, 2*1, 3*1, 4*1 or 1*2*3*4*1? (Both lists being natural numbers sequences starting in 1)
– Netwave
Mar 25 at 7:58
a=[1,2,3,.....40] len(a)=40 b=[1,2,....10] len(b)=10
– mashedpoteto
Mar 25 at 8:14
I want to make two lists,one by multiplying & other by dividing ->every first four element of a[ ] with the every element of b [ ] in loop. So something like : [1,2,3,4]/1 -> will be the first element of my new list. The second will be [5,6,7,8]/2 and so on.I hope this makes clear. The last element of the new set would be [37,38,39,40]/10. The above example was for division operator.I want to repeat the same thing with multiplication.
– mashedpoteto
Mar 25 at 8:19
|
show 2 more comments
Sample lists with what you have already tried would boost your chances of getting valid answers.
– DirtyBit
Mar 25 at 7:54
Can we see some code? Give it a shot.
– Aran-Fey
Mar 25 at 7:55
Multiply them how? 1*1, 2*1, 3*1, 4*1 or 1*2*3*4*1? (Both lists being natural numbers sequences starting in 1)
– Netwave
Mar 25 at 7:58
a=[1,2,3,.....40] len(a)=40 b=[1,2,....10] len(b)=10
– mashedpoteto
Mar 25 at 8:14
I want to make two lists,one by multiplying & other by dividing ->every first four element of a[ ] with the every element of b [ ] in loop. So something like : [1,2,3,4]/1 -> will be the first element of my new list. The second will be [5,6,7,8]/2 and so on.I hope this makes clear. The last element of the new set would be [37,38,39,40]/10. The above example was for division operator.I want to repeat the same thing with multiplication.
– mashedpoteto
Mar 25 at 8:19
Sample lists with what you have already tried would boost your chances of getting valid answers.
– DirtyBit
Mar 25 at 7:54
Sample lists with what you have already tried would boost your chances of getting valid answers.
– DirtyBit
Mar 25 at 7:54
Can we see some code? Give it a shot.
– Aran-Fey
Mar 25 at 7:55
Can we see some code? Give it a shot.
– Aran-Fey
Mar 25 at 7:55
Multiply them how? 1*1, 2*1, 3*1, 4*1 or 1*2*3*4*1? (Both lists being natural numbers sequences starting in 1)
– Netwave
Mar 25 at 7:58
Multiply them how? 1*1, 2*1, 3*1, 4*1 or 1*2*3*4*1? (Both lists being natural numbers sequences starting in 1)
– Netwave
Mar 25 at 7:58
a=[1,2,3,.....40] len(a)=40 b=[1,2,....10] len(b)=10
– mashedpoteto
Mar 25 at 8:14
a=[1,2,3,.....40] len(a)=40 b=[1,2,....10] len(b)=10
– mashedpoteto
Mar 25 at 8:14
I want to make two lists,one by multiplying & other by dividing ->every first four element of a[ ] with the every element of b [ ] in loop. So something like : [1,2,3,4]/1 -> will be the first element of my new list. The second will be [5,6,7,8]/2 and so on.I hope this makes clear. The last element of the new set would be [37,38,39,40]/10. The above example was for division operator.I want to repeat the same thing with multiplication.
– mashedpoteto
Mar 25 at 8:19
I want to make two lists,one by multiplying & other by dividing ->every first four element of a[ ] with the every element of b [ ] in loop. So something like : [1,2,3,4]/1 -> will be the first element of my new list. The second will be [5,6,7,8]/2 and so on.I hope this makes clear. The last element of the new set would be [37,38,39,40]/10. The above example was for division operator.I want to repeat the same thing with multiplication.
– mashedpoteto
Mar 25 at 8:19
|
show 2 more comments
2 Answers
2
active
oldest
votes
0
For e.g.
a = range(40)
b = range(10)
Simplest:
[x * b[i//4] for i, x in enumerate(a)]
More functional:
# from https://docs.python.org/3/library/itertools.html#itertools-recipes
from itertools import zip_longest
def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
args = [iter(iterable)] * n
return zip_longest(*args, fillvalue=fillvalue)
[a1 * b1 for a4, b1 in zip(grouper(a, 4), b) for a1 in a4]
answered Mar 25 at 7:58
AmadanAmadan
138k13152203
I just posted a sample comment above,I tried this out but I am unfortunately not getting the desired result.
– mashedpoteto
Mar 25 at 8:21
@mashedpoteto Looks like the desired result to me. If this isn't what you want, show us what is that you want and explain how this is different from that. (And don't do it in comments... edit it into your question.)
– Aran-Fey
Mar 25 at 8:27
@Aran-Fey edited it :) Can you take a look?
– mashedpoteto
Mar 25 at 8:53
Rolled back because the first solution was actually the correct one (yielding a list of 40 numbers, not a list 10 lists of 4); "Not getting the desired result" - you should be more specific how the "desired result" differs from the one obtained by my code.
– Amadan
Mar 25 at 8:56
@Amadan it worked.Initially the problem was a very simple error on how I wrote the code,which gave a division by zero error.Thank you :)
– mashedpoteto
Mar 26 at 12:14
add a comment |
0
You can do it using pure Python in one line:
newList = [active[j] / passive[i] for i in range(len(passive)) for j in range(i*4,i*4+4)]
for elem in newList:
print(elem)
This will print:
1.0454586587604597
0.9936967470945319
1.0078963662125922
1.0137739110579735
1.0366928664488493
1.0274097658218595
1.0399177333006342
1.040568497228071
1.020384882546816
1.0368647296698332
1.0230744951511204
1.0240563338877238
0.9836830066620091
1.0492516490722763
1.013507569945381
1.045643691807429
0.9824440408551517
1.0065553261670555
0.9834944282126284
0.982767218109524
0.9698119879004422
0.9690746274955083
0.9504097477092123
0.9750000553011796
0.9526234977470646
0.954087546649548
0.9548685224696527
0.9386092361191739
1.005980357871888
0.9894278559960512
0.9982460015709302
1.0227227411047006
0.9829530113857514
0.9616405454531767
0.9437321600631335
0.9532955903958973
0.9637894441999811
0.9706273811757119
0.9668057773255447
0.9430261366318299
Same goes for multiplying:
newList = [active[j] * passive[i] for i in range(len(passive)) for j in range(i*4,i*4+4)]
answered Mar 25 at 10:19
Vasilis G.Vasilis G.
4,3462925
add a comment |
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0
For e.g.
a = range(40)
b = range(10)
Simplest:
[x * b[i//4] for i, x in enumerate(a)]
More functional:
# from https://docs.python.org/3/library/itertools.html#itertools-recipes
from itertools import zip_longest
def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
args = [iter(iterable)] * n
return zip_longest(*args, fillvalue=fillvalue)
[a1 * b1 for a4, b1 in zip(grouper(a, 4), b) for a1 in a4]
answered Mar 25 at 7:58
AmadanAmadan
138k13152203
I just posted a sample comment above,I tried this out but I am unfortunately not getting the desired result.
– mashedpoteto
Mar 25 at 8:21
@mashedpoteto Looks like the desired result to me. If this isn't what you want, show us what is that you want and explain how this is different from that. (And don't do it in comments... edit it into your question.)
– Aran-Fey
Mar 25 at 8:27
@Aran-Fey edited it :) Can you take a look?
– mashedpoteto
Mar 25 at 8:53
Rolled back because the first solution was actually the correct one (yielding a list of 40 numbers, not a list 10 lists of 4); "Not getting the desired result" - you should be more specific how the "desired result" differs from the one obtained by my code.
– Amadan
Mar 25 at 8:56
@Amadan it worked.Initially the problem was a very simple error on how I wrote the code,which gave a division by zero error.Thank you :)
– mashedpoteto
Mar 26 at 12:14
add a comment |
0
For e.g.
a = range(40)
b = range(10)
Simplest:
[x * b[i//4] for i, x in enumerate(a)]
More functional:
# from https://docs.python.org/3/library/itertools.html#itertools-recipes
from itertools import zip_longest
def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
args = [iter(iterable)] * n
return zip_longest(*args, fillvalue=fillvalue)
[a1 * b1 for a4, b1 in zip(grouper(a, 4), b) for a1 in a4]
answered Mar 25 at 7:58
AmadanAmadan
138k13152203
I just posted a sample comment above,I tried this out but I am unfortunately not getting the desired result.
– mashedpoteto
Mar 25 at 8:21
@mashedpoteto Looks like the desired result to me. If this isn't what you want, show us what is that you want and explain how this is different from that. (And don't do it in comments... edit it into your question.)
– Aran-Fey
Mar 25 at 8:27
@Aran-Fey edited it :) Can you take a look?
– mashedpoteto
Mar 25 at 8:53
Rolled back because the first solution was actually the correct one (yielding a list of 40 numbers, not a list 10 lists of 4); "Not getting the desired result" - you should be more specific how the "desired result" differs from the one obtained by my code.
– Amadan
Mar 25 at 8:56
@Amadan it worked.Initially the problem was a very simple error on how I wrote the code,which gave a division by zero error.Thank you :)
– mashedpoteto
Mar 26 at 12:14
add a comment |
0
0
0
For e.g.
a = range(40)
b = range(10)
Simplest:
[x * b[i//4] for i, x in enumerate(a)]
More functional:
# from https://docs.python.org/3/library/itertools.html#itertools-recipes
from itertools import zip_longest
def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
args = [iter(iterable)] * n
return zip_longest(*args, fillvalue=fillvalue)
[a1 * b1 for a4, b1 in zip(grouper(a, 4), b) for a1 in a4]
answered Mar 25 at 7:58
AmadanAmadan
138k13152203
For e.g.
a = range(40)
b = range(10)
Simplest:
[x * b[i//4] for i, x in enumerate(a)]
More functional:
# from https://docs.python.org/3/library/itertools.html#itertools-recipes
from itertools import zip_longest
def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
args = [iter(iterable)] * n
return zip_longest(*args, fillvalue=fillvalue)
[a1 * b1 for a4, b1 in zip(grouper(a, 4), b) for a1 in a4]
answered Mar 25 at 7:58
AmadanAmadan
138k13152203
edited Mar 25 at 8:55
answered Mar 25 at 7:58
AmadanAmadan
138k13152203
answered Mar 25 at 7:58
AmadanAmadan
138k13152203
answered Mar 25 at 7:58
AmadanAmadan
138k13152203
138k13152203
I just posted a sample comment above,I tried this out but I am unfortunately not getting the desired result.
– mashedpoteto
Mar 25 at 8:21
@mashedpoteto Looks like the desired result to me. If this isn't what you want, show us what is that you want and explain how this is different from that. (And don't do it in comments... edit it into your question.)
– Aran-Fey
Mar 25 at 8:27
@Aran-Fey edited it :) Can you take a look?
– mashedpoteto
Mar 25 at 8:53
Rolled back because the first solution was actually the correct one (yielding a list of 40 numbers, not a list 10 lists of 4); "Not getting the desired result" - you should be more specific how the "desired result" differs from the one obtained by my code.
– Amadan
Mar 25 at 8:56
@Amadan it worked.Initially the problem was a very simple error on how I wrote the code,which gave a division by zero error.Thank you :)
– mashedpoteto
Mar 26 at 12:14
add a comment |
I just posted a sample comment above,I tried this out but I am unfortunately not getting the desired result.
– mashedpoteto
Mar 25 at 8:21
@mashedpoteto Looks like the desired result to me. If this isn't what you want, show us what is that you want and explain how this is different from that. (And don't do it in comments... edit it into your question.)
– Aran-Fey
Mar 25 at 8:27
@Aran-Fey edited it :) Can you take a look?
– mashedpoteto
Mar 25 at 8:53
Rolled back because the first solution was actually the correct one (yielding a list of 40 numbers, not a list 10 lists of 4); "Not getting the desired result" - you should be more specific how the "desired result" differs from the one obtained by my code.
– Amadan
Mar 25 at 8:56
@Amadan it worked.Initially the problem was a very simple error on how I wrote the code,which gave a division by zero error.Thank you :)
– mashedpoteto
Mar 26 at 12:14
I just posted a sample comment above,I tried this out but I am unfortunately not getting the desired result.
– mashedpoteto
Mar 25 at 8:21
I just posted a sample comment above,I tried this out but I am unfortunately not getting the desired result.
– mashedpoteto
Mar 25 at 8:21
@mashedpoteto Looks like the desired result to me. If this isn't what you want, show us what is that you want and explain how this is different from that. (And don't do it in comments... edit it into your question.)
– Aran-Fey
Mar 25 at 8:27
@mashedpoteto Looks like the desired result to me. If this isn't what you want, show us what is that you want and explain how this is different from that. (And don't do it in comments... edit it into your question.)
– Aran-Fey
Mar 25 at 8:27
@Aran-Fey edited it :) Can you take a look?
– mashedpoteto
Mar 25 at 8:53
@Aran-Fey edited it :) Can you take a look?
– mashedpoteto
Mar 25 at 8:53
Rolled back because the first solution was actually the correct one (yielding a list of 40 numbers, not a list 10 lists of 4); "Not getting the desired result" - you should be more specific how the "desired result" differs from the one obtained by my code.
– Amadan
Mar 25 at 8:56
Rolled back because the first solution was actually the correct one (yielding a list of 40 numbers, not a list 10 lists of 4); "Not getting the desired result" - you should be more specific how the "desired result" differs from the one obtained by my code.
– Amadan
Mar 25 at 8:56
@Amadan it worked.Initially the problem was a very simple error on how I wrote the code,which gave a division by zero error.Thank you :)
– mashedpoteto
Mar 26 at 12:14
@Amadan it worked.Initially the problem was a very simple error on how I wrote the code,which gave a division by zero error.Thank you :)
– mashedpoteto
Mar 26 at 12:14
add a comment |
0
You can do it using pure Python in one line:
newList = [active[j] / passive[i] for i in range(len(passive)) for j in range(i*4,i*4+4)]
for elem in newList:
print(elem)
This will print:
1.0454586587604597
0.9936967470945319
1.0078963662125922
1.0137739110579735
1.0366928664488493
1.0274097658218595
1.0399177333006342
1.040568497228071
1.020384882546816
1.0368647296698332
1.0230744951511204
1.0240563338877238
0.9836830066620091
1.0492516490722763
1.013507569945381
1.045643691807429
0.9824440408551517
1.0065553261670555
0.9834944282126284
0.982767218109524
0.9698119879004422
0.9690746274955083
0.9504097477092123
0.9750000553011796
0.9526234977470646
0.954087546649548
0.9548685224696527
0.9386092361191739
1.005980357871888
0.9894278559960512
0.9982460015709302
1.0227227411047006
0.9829530113857514
0.9616405454531767
0.9437321600631335
0.9532955903958973
0.9637894441999811
0.9706273811757119
0.9668057773255447
0.9430261366318299
Same goes for multiplying:
newList = [active[j] * passive[i] for i in range(len(passive)) for j in range(i*4,i*4+4)]
answered Mar 25 at 10:19
Vasilis G.Vasilis G.
4,3462925
add a comment |
0
You can do it using pure Python in one line:
newList = [active[j] / passive[i] for i in range(len(passive)) for j in range(i*4,i*4+4)]
for elem in newList:
print(elem)
This will print:
1.0454586587604597
0.9936967470945319
1.0078963662125922
1.0137739110579735
1.0366928664488493
1.0274097658218595
1.0399177333006342
1.040568497228071
1.020384882546816
1.0368647296698332
1.0230744951511204
1.0240563338877238
0.9836830066620091
1.0492516490722763
1.013507569945381
1.045643691807429
0.9824440408551517
1.0065553261670555
0.9834944282126284
0.982767218109524
0.9698119879004422
0.9690746274955083
0.9504097477092123
0.9750000553011796
0.9526234977470646
0.954087546649548
0.9548685224696527
0.9386092361191739
1.005980357871888
0.9894278559960512
0.9982460015709302
1.0227227411047006
0.9829530113857514
0.9616405454531767
0.9437321600631335
0.9532955903958973
0.9637894441999811
0.9706273811757119
0.9668057773255447
0.9430261366318299
Same goes for multiplying:
newList = [active[j] * passive[i] for i in range(len(passive)) for j in range(i*4,i*4+4)]
answered Mar 25 at 10:19
Vasilis G.Vasilis G.
4,3462925
add a comment |
0
0
0
You can do it using pure Python in one line:
newList = [active[j] / passive[i] for i in range(len(passive)) for j in range(i*4,i*4+4)]
for elem in newList:
print(elem)
This will print:
1.0454586587604597
0.9936967470945319
1.0078963662125922
1.0137739110579735
1.0366928664488493
1.0274097658218595
1.0399177333006342
1.040568497228071
1.020384882546816
1.0368647296698332
1.0230744951511204
1.0240563338877238
0.9836830066620091
1.0492516490722763
1.013507569945381
1.045643691807429
0.9824440408551517
1.0065553261670555
0.9834944282126284
0.982767218109524
0.9698119879004422
0.9690746274955083
0.9504097477092123
0.9750000553011796
0.9526234977470646
0.954087546649548
0.9548685224696527
0.9386092361191739
1.005980357871888
0.9894278559960512
0.9982460015709302
1.0227227411047006
0.9829530113857514
0.9616405454531767
0.9437321600631335
0.9532955903958973
0.9637894441999811
0.9706273811757119
0.9668057773255447
0.9430261366318299
Same goes for multiplying:
newList = [active[j] * passive[i] for i in range(len(passive)) for j in range(i*4,i*4+4)]
answered Mar 25 at 10:19
Vasilis G.Vasilis G.
4,3462925
You can do it using pure Python in one line:
newList = [active[j] / passive[i] for i in range(len(passive)) for j in range(i*4,i*4+4)]
for elem in newList:
print(elem)
This will print:
1.0454586587604597
0.9936967470945319
1.0078963662125922
1.0137739110579735
1.0366928664488493
1.0274097658218595
1.0399177333006342
1.040568497228071
1.020384882546816
1.0368647296698332
1.0230744951511204
1.0240563338877238
0.9836830066620091
1.0492516490722763
1.013507569945381
1.045643691807429
0.9824440408551517
1.0065553261670555
0.9834944282126284
0.982767218109524
0.9698119879004422
0.9690746274955083
0.9504097477092123
0.9750000553011796
0.9526234977470646
0.954087546649548
0.9548685224696527
0.9386092361191739
1.005980357871888
0.9894278559960512
0.9982460015709302
1.0227227411047006
0.9829530113857514
0.9616405454531767
0.9437321600631335
0.9532955903958973
0.9637894441999811
0.9706273811757119
0.9668057773255447
0.9430261366318299
Same goes for multiplying:
newList = [active[j] * passive[i] for i in range(len(passive)) for j in range(i*4,i*4+4)]
answered Mar 25 at 10:19
Vasilis G.Vasilis G.
4,3462925
answered Mar 25 at 10:19
Vasilis G.Vasilis G.
4,3462925
answered Mar 25 at 10:19
Vasilis G.Vasilis G.
4,3462925
answered Mar 25 at 10:19
Vasilis G.Vasilis G.
4,3462925
4,3462925
add a comment |
add a comment |
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Sample lists with what you have already tried would boost your chances of getting valid answers.
– DirtyBit
Mar 25 at 7:54
Can we see some code? Give it a shot.
– Aran-Fey
Mar 25 at 7:55
Multiply them how? 1*1, 2*1, 3*1, 4*1 or 1*2*3*4*1? (Both lists being natural numbers sequences starting in 1)
– Netwave
Mar 25 at 7:58
a=[1,2,3,.....40] len(a)=40 b=[1,2,....10] len(b)=10
– mashedpoteto
Mar 25 at 8:14
I want to make two lists,one by multiplying & other by dividing ->every first four element of a[ ] with the every element of b [ ] in loop. So something like : [1,2,3,4]/1 -> will be the first element of my new list. The second will be [5,6,7,8]/2 and so on.I hope this makes clear. The last element of the new set would be [37,38,39,40]/10. The above example was for division operator.I want to repeat the same thing with multiplication.
– mashedpoteto
Mar 25 at 8:19