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Unique identifier for each list of list in data frame

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-1

I have a list of list,

lst = [[2, 0, 1, 6, 7, 8], [4, 3, 5]] 

and I want to flatten the list and assign a unique id to each list in the list merged into a data.frame.

Desired output:

 value group
0 2 0
1 0 0
2 1 0
3 6 0
4 7 0
5 8 0
6 4 1
7 3 1
8 5 1

python

share|improve this question

asked Mar 26 at 18:42

VeddaVedda

2,52922 gold badges2222 silver badges5050 bronze badges

add a comment | 

-1

I have a list of list,

lst = [[2, 0, 1, 6, 7, 8], [4, 3, 5]] 

and I want to flatten the list and assign a unique id to each list in the list merged into a data.frame.

Desired output:

 value group
0 2 0
1 0 0
2 1 0
3 6 0
4 7 0
5 8 0
6 4 1
7 3 1
8 5 1

python

share|improve this question

asked Mar 26 at 18:42

VeddaVedda

2,52922 gold badges2222 silver badges5050 bronze badges

add a comment | 

I have a list of list,

lst = [[2, 0, 1, 6, 7, 8], [4, 3, 5]] 

and I want to flatten the list and assign a unique id to each list in the list merged into a data.frame.

Desired output:

 value group
0 2 0
1 0 0
2 1 0
3 6 0
4 7 0
5 8 0
6 4 1
7 3 1
8 5 1

python

share|improve this question

asked Mar 26 at 18:42

VeddaVedda

2,52922 gold badges2222 silver badges5050 bronze badges

lst = [[2, 0, 1, 6, 7, 8], [4, 3, 5]] 

 value group
0 2 0
1 0 0
2 1 0
3 6 0
4 7 0
5 8 0
6 4 1
7 3 1
8 5 1

share|improve this question

asked Mar 26 at 18:42

VeddaVedda

2,52922 gold badges2222 silver badges5050 bronze badges

share|improve this question

asked Mar 26 at 18:42

VeddaVedda

2,52922 gold badges2222 silver badges5050 bronze badges

asked Mar 26 at 18:42

VeddaVedda

2,52922 gold badges2222 silver badges5050 bronze badges

asked Mar 26 at 18:42

VeddaVedda

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4 Answers
4

active

oldest

votes

1

If you want a Pandas DataFrame:

import pandas as pd

lst = [[2, 0, 1, 6, 7, 8], [4, 3, 5]] 

final_list = []
for i, l in enumerate(lst):
 for num in l:
 final_list.append('value': num, 'group': i)

df = pd.DataFrame(final_list)

share|improve this answer

edited Mar 26 at 23:47

Luke Ning

11766 bronze badges

answered Mar 26 at 18:48

liamhawkinsliamhawkins

67611 gold badge66 silver badges2323 bronze badges

add a comment | 

2

You're going to need to do some fancy flattening:

flattened = [(item, index) for index, sublist in enumerate(lst) for item in sublist]
df = pd.DataFrame(flattened, columns=['value','group'])

share|improve this answer

answered Mar 26 at 18:48

ameame

27622 silver badges55 bronze badges

add a comment | 

1

you can use this code:

new_lst = []
for group in lst:
 for n in group:
 new_lst.append("group":lst.index(group),"value": n)

share|improve this answer

answered Mar 26 at 19:13

Ali HallajiAli Hallaji

1,0011212 silver badges2020 bronze badges

add a comment | 

0

You should try something before asking for desired output.

Looping through a list of list, whilst having a unique identifier, you may want to use the function enumerate that "gives the indexer" of the list.

for i,sub_list in enumerate(lst):
 identifier = i
 [(value,identifier) for value in sublist]
 ....

Hoping this will help

share|improve this answer

answered Mar 26 at 18:47

Born Tbe WastedBorn Tbe Wasted

5731313 bronze badges

add a comment | 

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1

If you want a Pandas DataFrame:

import pandas as pd

lst = [[2, 0, 1, 6, 7, 8], [4, 3, 5]] 

final_list = []
for i, l in enumerate(lst):
 for num in l:
 final_list.append('value': num, 'group': i)

df = pd.DataFrame(final_list)

share|improve this answer

edited Mar 26 at 23:47

Luke Ning

11766 bronze badges

answered Mar 26 at 18:48

liamhawkinsliamhawkins

67611 gold badge66 silver badges2323 bronze badges

add a comment | 

1

If you want a Pandas DataFrame:

import pandas as pd

lst = [[2, 0, 1, 6, 7, 8], [4, 3, 5]] 

final_list = []
for i, l in enumerate(lst):
 for num in l:
 final_list.append('value': num, 'group': i)

df = pd.DataFrame(final_list)

share|improve this answer

edited Mar 26 at 23:47

Luke Ning

11766 bronze badges

answered Mar 26 at 18:48

liamhawkinsliamhawkins

67611 gold badge66 silver badges2323 bronze badges

add a comment | 

If you want a Pandas DataFrame:

import pandas as pd

lst = [[2, 0, 1, 6, 7, 8], [4, 3, 5]] 

final_list = []
for i, l in enumerate(lst):
 for num in l:
 final_list.append('value': num, 'group': i)

df = pd.DataFrame(final_list)

share|improve this answer

edited Mar 26 at 23:47

Luke Ning

11766 bronze badges

answered Mar 26 at 18:48

liamhawkinsliamhawkins

67611 gold badge66 silver badges2323 bronze badges

import pandas as pd

lst = [[2, 0, 1, 6, 7, 8], [4, 3, 5]] 

final_list = []
for i, l in enumerate(lst):
 for num in l:
 final_list.append('value': num, 'group': i)

df = pd.DataFrame(final_list)

share|improve this answer

edited Mar 26 at 23:47

Luke Ning

11766 bronze badges

answered Mar 26 at 18:48

liamhawkinsliamhawkins

67611 gold badge66 silver badges2323 bronze badges

edited Mar 26 at 23:47

Luke Ning

11766 bronze badges

edited Mar 26 at 23:47

Luke Ning

11766 bronze badges

answered Mar 26 at 18:48

liamhawkinsliamhawkins

67611 gold badge66 silver badges2323 bronze badges

answered Mar 26 at 18:48

liamhawkinsliamhawkins

67611 gold badge66 silver badges2323 bronze badges

2

You're going to need to do some fancy flattening:

flattened = [(item, index) for index, sublist in enumerate(lst) for item in sublist]
df = pd.DataFrame(flattened, columns=['value','group'])

share|improve this answer

answered Mar 26 at 18:48

ameame

27622 silver badges55 bronze badges

add a comment | 

2

You're going to need to do some fancy flattening:

flattened = [(item, index) for index, sublist in enumerate(lst) for item in sublist]
df = pd.DataFrame(flattened, columns=['value','group'])

share|improve this answer

answered Mar 26 at 18:48

ameame

27622 silver badges55 bronze badges

add a comment | 

You're going to need to do some fancy flattening:

flattened = [(item, index) for index, sublist in enumerate(lst) for item in sublist]
df = pd.DataFrame(flattened, columns=['value','group'])

share|improve this answer

answered Mar 26 at 18:48

ameame

27622 silver badges55 bronze badges

flattened = [(item, index) for index, sublist in enumerate(lst) for item in sublist]
df = pd.DataFrame(flattened, columns=['value','group'])

share|improve this answer

answered Mar 26 at 18:48

ameame

27622 silver badges55 bronze badges

answered Mar 26 at 18:48

ameame

27622 silver badges55 bronze badges

answered Mar 26 at 18:48

ameame

27622 silver badges55 bronze badges

1

you can use this code:

new_lst = []
for group in lst:
 for n in group:
 new_lst.append("group":lst.index(group),"value": n)

share|improve this answer

answered Mar 26 at 19:13

Ali HallajiAli Hallaji

1,0011212 silver badges2020 bronze badges

add a comment | 

1

you can use this code:

new_lst = []
for group in lst:
 for n in group:
 new_lst.append("group":lst.index(group),"value": n)

share|improve this answer

answered Mar 26 at 19:13

Ali HallajiAli Hallaji

1,0011212 silver badges2020 bronze badges

add a comment | 

you can use this code:

new_lst = []
for group in lst:
 for n in group:
 new_lst.append("group":lst.index(group),"value": n)

share|improve this answer

answered Mar 26 at 19:13

Ali HallajiAli Hallaji

1,0011212 silver badges2020 bronze badges

new_lst = []
for group in lst:
 for n in group:
 new_lst.append("group":lst.index(group),"value": n)

share|improve this answer

answered Mar 26 at 19:13

Ali HallajiAli Hallaji

1,0011212 silver badges2020 bronze badges

answered Mar 26 at 19:13

Ali HallajiAli Hallaji

1,0011212 silver badges2020 bronze badges

answered Mar 26 at 19:13

Ali HallajiAli Hallaji

1,0011212 silver badges2020 bronze badges

0

You should try something before asking for desired output.

Looping through a list of list, whilst having a unique identifier, you may want to use the function enumerate that "gives the indexer" of the list.

for i,sub_list in enumerate(lst):
 identifier = i
 [(value,identifier) for value in sublist]
 ....

Hoping this will help

share|improve this answer

answered Mar 26 at 18:47

Born Tbe WastedBorn Tbe Wasted

5731313 bronze badges

add a comment | 

0

You should try something before asking for desired output.

Looping through a list of list, whilst having a unique identifier, you may want to use the function enumerate that "gives the indexer" of the list.

for i,sub_list in enumerate(lst):
 identifier = i
 [(value,identifier) for value in sublist]
 ....

Hoping this will help

share|improve this answer

answered Mar 26 at 18:47

Born Tbe WastedBorn Tbe Wasted

5731313 bronze badges

add a comment | 

You should try something before asking for desired output.

Looping through a list of list, whilst having a unique identifier, you may want to use the function enumerate that "gives the indexer" of the list.

for i,sub_list in enumerate(lst):
 identifier = i
 [(value,identifier) for value in sublist]
 ....

Hoping this will help

share|improve this answer

answered Mar 26 at 18:47

Born Tbe WastedBorn Tbe Wasted

5731313 bronze badges

for i,sub_list in enumerate(lst):
 identifier = i
 [(value,identifier) for value in sublist]
 ....

share|improve this answer

answered Mar 26 at 18:47

Born Tbe WastedBorn Tbe Wasted

5731313 bronze badges

answered Mar 26 at 18:47

Born Tbe WastedBorn Tbe Wasted

5731313 bronze badges

answered Mar 26 at 18:47

Born Tbe WastedBorn Tbe Wasted

5731313 bronze badges