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Inserting the element-wise mean of nested list into the same list
How to remove an element from a list by index?Find intersection of two nested lists?Getting the last element of a listHow do I get the number of elements in a list?Is there a simple way to delete a list element by value?Proper way to make HTML nested list?How to test if variable exists in nested list and append it to that list in pythonpython: list of lists to ordered dict and group by first elementcompare 2 2D lists and delete items from one list based on anotherIterating through database with nested lists
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Suppose there is a list of nested lists of floats
L = [[a,b,c],[e,f,g],[h,i,j]]
What kind of function can I define to iterate through the list once and insert the mean of elements of every consecutive list into the same list? I.e. I want to get
L1 = [[a,b,c],[(a+e)/2,(b+f)/2,(c+g)/2],[e,f,g],[(e+h)/2,(f+i)/2,(g+j)/2],[h,i,j]]
I know the function to get the element wise mean of two lists:
from operator import add
new_list = list(map(add,list1,list2))
J = [j/2 for j in new_list]
However inserting this list of mean values back into the same list while maintaining the proper index iteration through the old list proved challenging.
python list nested-lists
asked Mar 28 at 11:08
NetUser5y62NetUser5y62
1065 bronze badges
add a comment
|
Suppose there is a list of nested lists of floats
L = [[a,b,c],[e,f,g],[h,i,j]]
What kind of function can I define to iterate through the list once and insert the mean of elements of every consecutive list into the same list? I.e. I want to get
L1 = [[a,b,c],[(a+e)/2,(b+f)/2,(c+g)/2],[e,f,g],[(e+h)/2,(f+i)/2,(g+j)/2],[h,i,j]]
I know the function to get the element wise mean of two lists:
from operator import add
new_list = list(map(add,list1,list2))
J = [j/2 for j in new_list]
However inserting this list of mean values back into the same list while maintaining the proper index iteration through the old list proved challenging.
python list nested-lists
asked Mar 28 at 11:08
NetUser5y62NetUser5y62
1065 bronze badges
Just construct a new list, and mutate back into the original list if necessary.
– Andras Deak
Mar 28 at 11:20
add a comment
|
Suppose there is a list of nested lists of floats
L = [[a,b,c],[e,f,g],[h,i,j]]
What kind of function can I define to iterate through the list once and insert the mean of elements of every consecutive list into the same list? I.e. I want to get
L1 = [[a,b,c],[(a+e)/2,(b+f)/2,(c+g)/2],[e,f,g],[(e+h)/2,(f+i)/2,(g+j)/2],[h,i,j]]
I know the function to get the element wise mean of two lists:
from operator import add
new_list = list(map(add,list1,list2))
J = [j/2 for j in new_list]
However inserting this list of mean values back into the same list while maintaining the proper index iteration through the old list proved challenging.
python list nested-lists
asked Mar 28 at 11:08
NetUser5y62NetUser5y62
1065 bronze badges
Suppose there is a list of nested lists of floats
L = [[a,b,c],[e,f,g],[h,i,j]]
What kind of function can I define to iterate through the list once and insert the mean of elements of every consecutive list into the same list? I.e. I want to get
L1 = [[a,b,c],[(a+e)/2,(b+f)/2,(c+g)/2],[e,f,g],[(e+h)/2,(f+i)/2,(g+j)/2],[h,i,j]]
I know the function to get the element wise mean of two lists:
from operator import add
new_list = list(map(add,list1,list2))
J = [j/2 for j in new_list]
However inserting this list of mean values back into the same list while maintaining the proper index iteration through the old list proved challenging.
python list nested-lists
python list nested-lists
asked Mar 28 at 11:08
NetUser5y62NetUser5y62
1065 bronze badges
asked Mar 28 at 11:08
NetUser5y62NetUser5y62
1065 bronze badges
asked Mar 28 at 11:08
NetUser5y62NetUser5y62
1065 bronze badges
asked Mar 28 at 11:08
NetUser5y62NetUser5y62
1065 bronze badges
asked Mar 28 at 11:08
NetUser5y62NetUser5y62
1065 bronze badges
1065 bronze badges
Just construct a new list, and mutate back into the original list if necessary.
– Andras Deak
Mar 28 at 11:20
add a comment
|
Just construct a new list, and mutate back into the original list if necessary.
– Andras Deak
Mar 28 at 11:20
Just construct a new list, and mutate back into the original list if necessary.
– Andras Deak
Mar 28 at 11:20
Just construct a new list, and mutate back into the original list if necessary.
– Andras Deak
Mar 28 at 11:20
add a comment
|
1 Answer
1
active
oldest
votes
There are two cases:
- You don't care if the resulting list is the same list:
new_list = []
for i in range(len(L)-1):
new_list.append(L[i])
new_list.append(list(map(lambda x: sum(x)/len(x), zip(L[i],L[i+1]))))
new_list.append(L[-1])
- You want the changes to be done in-place:
i=0
while i < len(L)-1:
new_elem = list(map(lambda x: sum(x)/len(x), zip(L[i],L[i+1])))
L.insert(i+1, new_elem)
i += 2
EDIT: If you're using python 3.4 or above, instead of lambda x: sum(x)/len(x) you can use mean(x) (from the package statistics).
answered Mar 28 at 11:23
BogsanBogsan
3612 silver badges7 bronze badges
1
thank you, for number 2 there is a small typo: i < len(L) - 1
– NetUser5y62
Mar 28 at 11:34
avoid usinglambdaand just use the builtinmeanfrom the stdlib in thestatisticslibrary
– aws_apprentice
Mar 28 at 11:42
add a comment
|
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
There are two cases:
- You don't care if the resulting list is the same list:
new_list = []
for i in range(len(L)-1):
new_list.append(L[i])
new_list.append(list(map(lambda x: sum(x)/len(x), zip(L[i],L[i+1]))))
new_list.append(L[-1])
- You want the changes to be done in-place:
i=0
while i < len(L)-1:
new_elem = list(map(lambda x: sum(x)/len(x), zip(L[i],L[i+1])))
L.insert(i+1, new_elem)
i += 2
EDIT: If you're using python 3.4 or above, instead of lambda x: sum(x)/len(x) you can use mean(x) (from the package statistics).
answered Mar 28 at 11:23
BogsanBogsan
3612 silver badges7 bronze badges
1
thank you, for number 2 there is a small typo: i < len(L) - 1
– NetUser5y62
Mar 28 at 11:34
avoid usinglambdaand just use the builtinmeanfrom the stdlib in thestatisticslibrary
– aws_apprentice
Mar 28 at 11:42
add a comment
|
There are two cases:
- You don't care if the resulting list is the same list:
new_list = []
for i in range(len(L)-1):
new_list.append(L[i])
new_list.append(list(map(lambda x: sum(x)/len(x), zip(L[i],L[i+1]))))
new_list.append(L[-1])
- You want the changes to be done in-place:
i=0
while i < len(L)-1:
new_elem = list(map(lambda x: sum(x)/len(x), zip(L[i],L[i+1])))
L.insert(i+1, new_elem)
i += 2
EDIT: If you're using python 3.4 or above, instead of lambda x: sum(x)/len(x) you can use mean(x) (from the package statistics).
answered Mar 28 at 11:23
BogsanBogsan
3612 silver badges7 bronze badges
1
thank you, for number 2 there is a small typo: i < len(L) - 1
– NetUser5y62
Mar 28 at 11:34
avoid usinglambdaand just use the builtinmeanfrom the stdlib in thestatisticslibrary
– aws_apprentice
Mar 28 at 11:42
add a comment
|
There are two cases:
- You don't care if the resulting list is the same list:
new_list = []
for i in range(len(L)-1):
new_list.append(L[i])
new_list.append(list(map(lambda x: sum(x)/len(x), zip(L[i],L[i+1]))))
new_list.append(L[-1])
- You want the changes to be done in-place:
i=0
while i < len(L)-1:
new_elem = list(map(lambda x: sum(x)/len(x), zip(L[i],L[i+1])))
L.insert(i+1, new_elem)
i += 2
EDIT: If you're using python 3.4 or above, instead of lambda x: sum(x)/len(x) you can use mean(x) (from the package statistics).
answered Mar 28 at 11:23
BogsanBogsan
3612 silver badges7 bronze badges
There are two cases:
- You don't care if the resulting list is the same list:
new_list = []
for i in range(len(L)-1):
new_list.append(L[i])
new_list.append(list(map(lambda x: sum(x)/len(x), zip(L[i],L[i+1]))))
new_list.append(L[-1])
- You want the changes to be done in-place:
i=0
while i < len(L)-1:
new_elem = list(map(lambda x: sum(x)/len(x), zip(L[i],L[i+1])))
L.insert(i+1, new_elem)
i += 2
EDIT: If you're using python 3.4 or above, instead of lambda x: sum(x)/len(x) you can use mean(x) (from the package statistics).
answered Mar 28 at 11:23
BogsanBogsan
3612 silver badges7 bronze badges
edited Mar 28 at 11:50
answered Mar 28 at 11:23
BogsanBogsan
3612 silver badges7 bronze badges
answered Mar 28 at 11:23
BogsanBogsan
3612 silver badges7 bronze badges
answered Mar 28 at 11:23
BogsanBogsan
3612 silver badges7 bronze badges
3612 silver badges7 bronze badges
1
thank you, for number 2 there is a small typo: i < len(L) - 1
– NetUser5y62
Mar 28 at 11:34
avoid usinglambdaand just use the builtinmeanfrom the stdlib in thestatisticslibrary
– aws_apprentice
Mar 28 at 11:42
add a comment
|
1
thank you, for number 2 there is a small typo: i < len(L) - 1
– NetUser5y62
Mar 28 at 11:34
avoid usinglambdaand just use the builtinmeanfrom the stdlib in thestatisticslibrary
– aws_apprentice
Mar 28 at 11:42
1
1
thank you, for number 2 there is a small typo: i < len(L) - 1
– NetUser5y62
Mar 28 at 11:34
thank you, for number 2 there is a small typo: i < len(L) - 1
– NetUser5y62
Mar 28 at 11:34
avoid using
lambda and just use the builtin mean from the stdlib in the statistics library– aws_apprentice
Mar 28 at 11:42
avoid using
lambda and just use the builtin mean from the stdlib in the statistics library– aws_apprentice
Mar 28 at 11:42
add a comment
|
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Just construct a new list, and mutate back into the original list if necessary.
– Andras Deak
Mar 28 at 11:20