Pandas: compare list objects in Series

up vote
6
down vote

favorite

In my dataframe a column is made up of lists, for example:

df = pd.DataFrame('A':[[1,2],[2,4],[3,1]])

I need to find out the location of list [1,2] in this dataframe. I tried:

df.loc[df['A'] == [1,2]]

and

df.loc[df['A'] == [[1,2]]]

but failed totally. The comparison seems very simple but that just doesn't work. Am I missing something here?

edited 1 hour ago

Seanny123

2,18833261

asked 2 hours ago

Shiang Hoo

311

New contributor

add a commentÂ |Â

up vote
6
down vote

favorite

In my dataframe a column is made up of lists, for example:

df = pd.DataFrame('A':[[1,2],[2,4],[3,1]])

I need to find out the location of list [1,2] in this dataframe. I tried:

df.loc[df['A'] == [1,2]]

and

df.loc[df['A'] == [[1,2]]]

but failed totally. The comparison seems very simple but that just doesn't work. Am I missing something here?

edited 1 hour ago

Seanny123

2,18833261

asked 2 hours ago

Shiang Hoo

311

New contributor

add a commentÂ |Â

up vote
6
down vote

favorite

In my dataframe a column is made up of lists, for example:

df = pd.DataFrame('A':[[1,2],[2,4],[3,1]])

I need to find out the location of list [1,2] in this dataframe. I tried:

df.loc[df['A'] == [1,2]]

and

df.loc[df['A'] == [[1,2]]]

but failed totally. The comparison seems very simple but that just doesn't work. Am I missing something here?

edited 1 hour ago

Seanny123

2,18833261

asked 2 hours ago

Shiang Hoo

311

New contributor

In my dataframe a column is made up of lists, for example:

df = pd.DataFrame('A':[[1,2],[2,4],[3,1]])

I need to find out the location of list [1,2] in this dataframe. I tried:

df.loc[df['A'] == [1,2]]

and

df.loc[df['A'] == [[1,2]]]

but failed totally. The comparison seems very simple but that just doesn't work. Am I missing something here?

python pandas

edited 1 hour ago

Seanny123

2,18833261

asked 2 hours ago

Shiang Hoo

311

New contributor

edited 1 hour ago

Seanny123

2,18833261

asked 2 hours ago

Shiang Hoo

311

New contributor

edited 1 hour ago

Seanny123

2,18833261

edited 1 hour ago

Seanny123

2,18833261

edited 1 hour ago

Seanny123

2,18833261

asked 2 hours ago

Shiang Hoo

311

New contributor

asked 2 hours ago

Shiang Hoo

311

asked 2 hours ago

Shiang Hoo

311

New contributor

Shiang Hoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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4 Answers
4

active

oldest

votes

up vote
6
down vote

Do not using list in cell , it create a lot of problem for pandas .If you do need a object columns , using tuple

df.A.map(tuple).isin([(1,2)])
Out[293]: 
0 True
1 False
2 False
Name: A, dtype: bool
#df[df.A.map(tuple).isin([(1,2)])]

answered 2 hours ago

W-B

89.1k72653

add a commentÂ |Â

up vote
5
down vote

With Numpy arrays

df.assign(B=(np.array(df.A.tolist()) == [1, 2]).all(1))

 A B
0 [1, 2] True
1 [2, 4] False
2 [3, 1] False

answered 1 hour ago

piRSquared

146k21128260

2

This should be the accepted solution! [Or, if possible, just expanding the series of lists to 2 series.]
â€“Â jpp
1 hour ago

Won't this run into issues if the lists are differently sized, though perhaps that's outside of the scope of this example.
â€“Â ALollz
21 mins ago

@ALollz yes and yes
â€“Â piRSquared
8 mins ago

add a commentÂ |Â

up vote
3
down vote

You can use apply and compare as:

df['A'].apply(lambda x: x==[1,2])

0 True
1 False
2 False
Name: A, dtype: bool

print(df[df['A'].apply(lambda x: x==[1,2])])

 A
0 [1, 2]

answered 2 hours ago

Sandeep Kadapa

3,524424

add a commentÂ |Â

up vote
2
down vote

Little messy solution using numpy

df.A.apply(lambda x: (np.array(x) == np.array([1,2])).all())

0 True
1 False
2 False

answered 1 hour ago

Vaishali

16k3927

add a commentÂ |Â

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

up vote
6
down vote

Do not using list in cell , it create a lot of problem for pandas .If you do need a object columns , using tuple

df.A.map(tuple).isin([(1,2)])
Out[293]: 
0 True
1 False
2 False
Name: A, dtype: bool
#df[df.A.map(tuple).isin([(1,2)])]

answered 2 hours ago

W-B

89.1k72653

add a commentÂ |Â

up vote
6
down vote

Do not using list in cell , it create a lot of problem for pandas .If you do need a object columns , using tuple

df.A.map(tuple).isin([(1,2)])
Out[293]: 
0 True
1 False
2 False
Name: A, dtype: bool
#df[df.A.map(tuple).isin([(1,2)])]

answered 2 hours ago

W-B

89.1k72653

add a commentÂ |Â

up vote
6
down vote

Do not using list in cell , it create a lot of problem for pandas .If you do need a object columns , using tuple

df.A.map(tuple).isin([(1,2)])
Out[293]: 
0 True
1 False
2 False
Name: A, dtype: bool
#df[df.A.map(tuple).isin([(1,2)])]

answered 2 hours ago

W-B

89.1k72653

Do not using list in cell , it create a lot of problem for pandas .If you do need a object columns , using tuple

df.A.map(tuple).isin([(1,2)])
Out[293]: 
0 True
1 False
2 False
Name: A, dtype: bool
#df[df.A.map(tuple).isin([(1,2)])]

answered 2 hours ago

W-B

89.1k72653

answered 2 hours ago

W-B

89.1k72653

answered 2 hours ago

W-B

89.1k72653

answered 2 hours ago

W-B

89.1k72653

add a commentÂ |Â

up vote
5
down vote

With Numpy arrays

df.assign(B=(np.array(df.A.tolist()) == [1, 2]).all(1))

 A B
0 [1, 2] True
1 [2, 4] False
2 [3, 1] False

answered 1 hour ago

piRSquared

146k21128260

2

This should be the accepted solution! [Or, if possible, just expanding the series of lists to 2 series.]
â€“Â jpp
1 hour ago

Won't this run into issues if the lists are differently sized, though perhaps that's outside of the scope of this example.
â€“Â ALollz
21 mins ago

@ALollz yes and yes
â€“Â piRSquared
8 mins ago

add a commentÂ |Â

up vote
5
down vote

With Numpy arrays

df.assign(B=(np.array(df.A.tolist()) == [1, 2]).all(1))

 A B
0 [1, 2] True
1 [2, 4] False
2 [3, 1] False

answered 1 hour ago

piRSquared

146k21128260

2

This should be the accepted solution! [Or, if possible, just expanding the series of lists to 2 series.]
â€“Â jpp
1 hour ago

Won't this run into issues if the lists are differently sized, though perhaps that's outside of the scope of this example.
â€“Â ALollz
21 mins ago

@ALollz yes and yes
â€“Â piRSquared
8 mins ago

add a commentÂ |Â

up vote
5
down vote

With Numpy arrays

df.assign(B=(np.array(df.A.tolist()) == [1, 2]).all(1))

 A B
0 [1, 2] True
1 [2, 4] False
2 [3, 1] False

answered 1 hour ago

piRSquared

146k21128260

With Numpy arrays

df.assign(B=(np.array(df.A.tolist()) == [1, 2]).all(1))

 A B
0 [1, 2] True
1 [2, 4] False
2 [3, 1] False

answered 1 hour ago

piRSquared

146k21128260

answered 1 hour ago

piRSquared

146k21128260

answered 1 hour ago

piRSquared

146k21128260

answered 1 hour ago

piRSquared

146k21128260

2

This should be the accepted solution! [Or, if possible, just expanding the series of lists to 2 series.]
â€“Â jpp
1 hour ago

Won't this run into issues if the lists are differently sized, though perhaps that's outside of the scope of this example.
â€“Â ALollz
21 mins ago

@ALollz yes and yes
â€“Â piRSquared
8 mins ago

add a commentÂ |Â

2

This should be the accepted solution! [Or, if possible, just expanding the series of lists to 2 series.]
â€“Â jpp
1 hour ago

Won't this run into issues if the lists are differently sized, though perhaps that's outside of the scope of this example.
â€“Â ALollz
21 mins ago

@ALollz yes and yes
â€“Â piRSquared
8 mins ago

This should be the accepted solution! [Or, if possible, just expanding the series of lists to 2 series.]
â€“Â jpp
1 hour ago

Won't this run into issues if the lists are differently sized, though perhaps that's outside of the scope of this example.
â€“Â ALollz
21 mins ago

@ALollz yes and yes
â€“Â piRSquared
8 mins ago

add a commentÂ |Â

up vote
3
down vote

You can use apply and compare as:

df['A'].apply(lambda x: x==[1,2])

0 True
1 False
2 False
Name: A, dtype: bool

print(df[df['A'].apply(lambda x: x==[1,2])])

 A
0 [1, 2]

answered 2 hours ago

Sandeep Kadapa

3,524424

add a commentÂ |Â

up vote
3
down vote

You can use apply and compare as:

df['A'].apply(lambda x: x==[1,2])

0 True
1 False
2 False
Name: A, dtype: bool

print(df[df['A'].apply(lambda x: x==[1,2])])

 A
0 [1, 2]

answered 2 hours ago

Sandeep Kadapa

3,524424

add a commentÂ |Â

up vote
3
down vote

You can use apply and compare as:

df['A'].apply(lambda x: x==[1,2])

0 True
1 False
2 False
Name: A, dtype: bool

print(df[df['A'].apply(lambda x: x==[1,2])])

 A
0 [1, 2]

answered 2 hours ago

Sandeep Kadapa

3,524424

You can use apply and compare as:

df['A'].apply(lambda x: x==[1,2])

0 True
1 False
2 False
Name: A, dtype: bool

print(df[df['A'].apply(lambda x: x==[1,2])])

 A
0 [1, 2]

answered 2 hours ago

Sandeep Kadapa

3,524424

answered 2 hours ago

Sandeep Kadapa

3,524424

answered 2 hours ago

Sandeep Kadapa

3,524424

answered 2 hours ago

Sandeep Kadapa

3,524424

add a commentÂ |Â

up vote
2
down vote

Little messy solution using numpy

df.A.apply(lambda x: (np.array(x) == np.array([1,2])).all())

0 True
1 False
2 False

answered 1 hour ago

Vaishali

16k3927

add a commentÂ |Â

up vote
2
down vote

Little messy solution using numpy

df.A.apply(lambda x: (np.array(x) == np.array([1,2])).all())

0 True
1 False
2 False

answered 1 hour ago

Vaishali

16k3927

add a commentÂ |Â

up vote
2
down vote

Little messy solution using numpy

df.A.apply(lambda x: (np.array(x) == np.array([1,2])).all())

0 True
1 False
2 False

answered 1 hour ago

Vaishali

16k3927

Little messy solution using numpy

df.A.apply(lambda x: (np.array(x) == np.array([1,2])).all())

0 True
1 False
2 False

answered 1 hour ago

Vaishali

16k3927

answered 1 hour ago

Vaishali

16k3927

answered 1 hour ago

Vaishali

16k3927

answered 1 hour ago

Vaishali

16k3927

add a commentÂ |Â

Shiang Hoo is a new contributor. Be nice, and check out our Code of Conduct.

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