Pandas: compare list objects in Series
10
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 Nov 1 '18 at 14:25
Seanny123
2,30743364
asked Nov 1 '18 at 13:53
Shiang HooShiang Hoo
794
add a comment |
10
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 Nov 1 '18 at 14:25
Seanny123
2,30743364
asked Nov 1 '18 at 13:53
Shiang HooShiang Hoo
794
The only thing you're "missing" is that data frames aren't really great for storing lists. Any reason you don't want two separate columns?
– BallpointBen
Nov 1 '18 at 21:18
@BallpointBen Thanks for your attention, I've posted a new question to explain the whole question. stackoverflow.com/questions/53115592/…
– Shiang Hoo
Nov 2 '18 at 9:11
@Luuklag This may be a duplicate, but I don't believe it's a duplicate of the target you suggest. That one seems to be trying to filter based on whether multiple columns are equal to particular values. This one is trying to check if the list is equal to a single column's value, which has a very different answer.
– jpmc26
Nov 13 '18 at 22:19
Feel free to suggest a more appropriate target.
– Luuklag
Nov 13 '18 at 23:05
@Luuklag, I posted the two questions because I don't think they are the same. As jpmc described, they are connected but also very different. This post is actually the varietas of that one: I tried stupid things to solve that one and based on the stupid thing I posted this one. But this one still has its distinct value. Can you please remove the duplicate target?
– Shiang Hoo
Nov 19 '18 at 2:56
add a comment |
10
10
10
2
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 Nov 1 '18 at 14:25
Seanny123
2,30743364
asked Nov 1 '18 at 13:53
Shiang HooShiang Hoo
794
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
python pandas
edited Nov 1 '18 at 14:25
Seanny123
2,30743364
asked Nov 1 '18 at 13:53
Shiang HooShiang Hoo
794
edited Nov 1 '18 at 14:25
Seanny123
2,30743364
asked Nov 1 '18 at 13:53
Shiang HooShiang Hoo
794
edited Nov 1 '18 at 14:25
Seanny123
2,30743364
edited Nov 1 '18 at 14:25
Seanny123
2,30743364
edited Nov 1 '18 at 14:25
Seanny123
2,30743364
2,30743364
asked Nov 1 '18 at 13:53
Shiang HooShiang Hoo
794
asked Nov 1 '18 at 13:53
Shiang HooShiang Hoo
794
asked Nov 1 '18 at 13:53
Shiang HooShiang Hoo
794
794
The only thing you're "missing" is that data frames aren't really great for storing lists. Any reason you don't want two separate columns?
– BallpointBen
Nov 1 '18 at 21:18
@BallpointBen Thanks for your attention, I've posted a new question to explain the whole question. stackoverflow.com/questions/53115592/…
– Shiang Hoo
Nov 2 '18 at 9:11
@Luuklag This may be a duplicate, but I don't believe it's a duplicate of the target you suggest. That one seems to be trying to filter based on whether multiple columns are equal to particular values. This one is trying to check if the list is equal to a single column's value, which has a very different answer.
– jpmc26
Nov 13 '18 at 22:19
Feel free to suggest a more appropriate target.
– Luuklag
Nov 13 '18 at 23:05
@Luuklag, I posted the two questions because I don't think they are the same. As jpmc described, they are connected but also very different. This post is actually the varietas of that one: I tried stupid things to solve that one and based on the stupid thing I posted this one. But this one still has its distinct value. Can you please remove the duplicate target?
– Shiang Hoo
Nov 19 '18 at 2:56
add a comment |
The only thing you're "missing" is that data frames aren't really great for storing lists. Any reason you don't want two separate columns?
– BallpointBen
Nov 1 '18 at 21:18
@BallpointBen Thanks for your attention, I've posted a new question to explain the whole question. stackoverflow.com/questions/53115592/…
– Shiang Hoo
Nov 2 '18 at 9:11
@Luuklag This may be a duplicate, but I don't believe it's a duplicate of the target you suggest. That one seems to be trying to filter based on whether multiple columns are equal to particular values. This one is trying to check if the list is equal to a single column's value, which has a very different answer.
– jpmc26
Nov 13 '18 at 22:19
Feel free to suggest a more appropriate target.
– Luuklag
Nov 13 '18 at 23:05
@Luuklag, I posted the two questions because I don't think they are the same. As jpmc described, they are connected but also very different. This post is actually the varietas of that one: I tried stupid things to solve that one and based on the stupid thing I posted this one. But this one still has its distinct value. Can you please remove the duplicate target?
– Shiang Hoo
Nov 19 '18 at 2:56
The only thing you're "missing" is that data frames aren't really great for storing lists. Any reason you don't want two separate columns?
– BallpointBen
Nov 1 '18 at 21:18
The only thing you're "missing" is that data frames aren't really great for storing lists. Any reason you don't want two separate columns?
– BallpointBen
Nov 1 '18 at 21:18
@BallpointBen Thanks for your attention, I've posted a new question to explain the whole question. stackoverflow.com/questions/53115592/…
– Shiang Hoo
Nov 2 '18 at 9:11
@BallpointBen Thanks for your attention, I've posted a new question to explain the whole question. stackoverflow.com/questions/53115592/…
– Shiang Hoo
Nov 2 '18 at 9:11
@Luuklag This may be a duplicate, but I don't believe it's a duplicate of the target you suggest. That one seems to be trying to filter based on whether multiple columns are equal to particular values. This one is trying to check if the list is equal to a single column's value, which has a very different answer.
– jpmc26
Nov 13 '18 at 22:19
@Luuklag This may be a duplicate, but I don't believe it's a duplicate of the target you suggest. That one seems to be trying to filter based on whether multiple columns are equal to particular values. This one is trying to check if the list is equal to a single column's value, which has a very different answer.
– jpmc26
Nov 13 '18 at 22:19
Feel free to suggest a more appropriate target.
– Luuklag
Nov 13 '18 at 23:05
Feel free to suggest a more appropriate target.
– Luuklag
Nov 13 '18 at 23:05
@Luuklag, I posted the two questions because I don't think they are the same. As jpmc described, they are connected but also very different. This post is actually the varietas of that one: I tried stupid things to solve that one and based on the stupid thing I posted this one. But this one still has its distinct value. Can you please remove the duplicate target?
– Shiang Hoo
Nov 19 '18 at 2:56
@Luuklag, I posted the two questions because I don't think they are the same. As jpmc described, they are connected but also very different. This post is actually the varietas of that one: I tried stupid things to solve that one and based on the stupid thing I posted this one. But this one still has its distinct value. Can you please remove the duplicate target?
– Shiang Hoo
Nov 19 '18 at 2:56
add a comment |
5 Answers
5
active
oldest
votes
12
Do not use list in cell, it creates a lot of problem for pandas. If you do need an object column, 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)])]
edited Nov 25 '18 at 19:42
anothermh
3,21331531
answered Nov 1 '18 at 13:56
W-BW-B
104k73165
add a comment |
10
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 Nov 1 '18 at 13:56
Sandeep KadapaSandeep Kadapa
6,277429
add a comment |
9
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 Nov 1 '18 at 14:34
piRSquaredpiRSquared
153k22144287
2
This should be the accepted solution! [Or, if possible, just expanding the series of lists to 2 series.]
– jpp
Nov 1 '18 at 14:43
Won't this run into issues if the lists are differently sized, though perhaps that's outside of the scope of this example.
– ALollz
Nov 1 '18 at 15:40
2
@ALollz yes and yes
– piRSquared
Nov 1 '18 at 15:53
Nice! My only concern is, this solution converts datatype twice, what if my dataframe is very big, will this conversion cost more time?
– Shiang Hoo
Nov 2 '18 at 2:41
add a comment |
6
Using numpy
df.A.apply(lambda x: (np.array(x) == np.array([1,2])).all())
0 True
1 False
2 False
answered Nov 1 '18 at 14:32
VaishaliVaishali
18.6k31030
add a comment |
0
Or:
df['A'].apply(([1,2]).__eq__)
Then:
df[df['A'].apply(([1,2]).__eq__)]
answered Nov 9 '18 at 4:18
U9-ForwardU9-Forward
14k21337
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
12
Do not use list in cell, it creates a lot of problem for pandas. If you do need an object column, 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)])]
edited Nov 25 '18 at 19:42
anothermh
3,21331531
answered Nov 1 '18 at 13:56
W-BW-B
104k73165
add a comment |
12
Do not use list in cell, it creates a lot of problem for pandas. If you do need an object column, 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)])]
edited Nov 25 '18 at 19:42
anothermh
3,21331531
answered Nov 1 '18 at 13:56
W-BW-B
104k73165
add a comment |
12
12
12
Do not use list in cell, it creates a lot of problem for pandas. If you do need an object column, 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)])]
edited Nov 25 '18 at 19:42
anothermh
3,21331531
answered Nov 1 '18 at 13:56
W-BW-B
104k73165
Do not use list in cell, it creates a lot of problem for pandas. If you do need an object column, 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)])]
edited Nov 25 '18 at 19:42
anothermh
3,21331531
answered Nov 1 '18 at 13:56
W-BW-B
104k73165
edited Nov 25 '18 at 19:42
anothermh
3,21331531
edited Nov 25 '18 at 19:42
anothermh
3,21331531
edited Nov 25 '18 at 19:42
anothermh
3,21331531
3,21331531
answered Nov 1 '18 at 13:56
W-BW-B
104k73165
answered Nov 1 '18 at 13:56
W-BW-B
104k73165
answered Nov 1 '18 at 13:56
W-BW-B
104k73165
104k73165
add a comment |
add a comment |
10
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 Nov 1 '18 at 13:56
Sandeep KadapaSandeep Kadapa
6,277429
add a comment |
10
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 Nov 1 '18 at 13:56
Sandeep KadapaSandeep Kadapa
6,277429
add a comment |
10
10
10
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 Nov 1 '18 at 13:56
Sandeep KadapaSandeep Kadapa
6,277429
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 Nov 1 '18 at 13:56
Sandeep KadapaSandeep Kadapa
6,277429
answered Nov 1 '18 at 13:56
Sandeep KadapaSandeep Kadapa
6,277429
answered Nov 1 '18 at 13:56
Sandeep KadapaSandeep Kadapa
6,277429
answered Nov 1 '18 at 13:56
Sandeep KadapaSandeep Kadapa
6,277429
6,277429
add a comment |
add a comment |
9
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 Nov 1 '18 at 14:34
piRSquaredpiRSquared
153k22144287
2
This should be the accepted solution! [Or, if possible, just expanding the series of lists to 2 series.]
– jpp
Nov 1 '18 at 14:43
Won't this run into issues if the lists are differently sized, though perhaps that's outside of the scope of this example.
– ALollz
Nov 1 '18 at 15:40
2
@ALollz yes and yes
– piRSquared
Nov 1 '18 at 15:53
Nice! My only concern is, this solution converts datatype twice, what if my dataframe is very big, will this conversion cost more time?
– Shiang Hoo
Nov 2 '18 at 2:41
add a comment |
9
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 Nov 1 '18 at 14:34
piRSquaredpiRSquared
153k22144287
2
This should be the accepted solution! [Or, if possible, just expanding the series of lists to 2 series.]
– jpp
Nov 1 '18 at 14:43
Won't this run into issues if the lists are differently sized, though perhaps that's outside of the scope of this example.
– ALollz
Nov 1 '18 at 15:40
2
@ALollz yes and yes
– piRSquared
Nov 1 '18 at 15:53
Nice! My only concern is, this solution converts datatype twice, what if my dataframe is very big, will this conversion cost more time?
– Shiang Hoo
Nov 2 '18 at 2:41
add a comment |
9
9
9
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 Nov 1 '18 at 14:34
piRSquaredpiRSquared
153k22144287
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 Nov 1 '18 at 14:34
piRSquaredpiRSquared
153k22144287
answered Nov 1 '18 at 14:34
piRSquaredpiRSquared
153k22144287
answered Nov 1 '18 at 14:34
piRSquaredpiRSquared
153k22144287
answered Nov 1 '18 at 14:34
piRSquaredpiRSquared
153k22144287
153k22144287
2
This should be the accepted solution! [Or, if possible, just expanding the series of lists to 2 series.]
– jpp
Nov 1 '18 at 14:43
Won't this run into issues if the lists are differently sized, though perhaps that's outside of the scope of this example.
– ALollz
Nov 1 '18 at 15:40
2
@ALollz yes and yes
– piRSquared
Nov 1 '18 at 15:53
Nice! My only concern is, this solution converts datatype twice, what if my dataframe is very big, will this conversion cost more time?
– Shiang Hoo
Nov 2 '18 at 2:41
add a comment |
2
This should be the accepted solution! [Or, if possible, just expanding the series of lists to 2 series.]
– jpp
Nov 1 '18 at 14:43
Won't this run into issues if the lists are differently sized, though perhaps that's outside of the scope of this example.
– ALollz
Nov 1 '18 at 15:40
2
@ALollz yes and yes
– piRSquared
Nov 1 '18 at 15:53
Nice! My only concern is, this solution converts datatype twice, what if my dataframe is very big, will this conversion cost more time?
– Shiang Hoo
Nov 2 '18 at 2:41
2
2
This should be the accepted solution! [Or, if possible, just expanding the series of lists to 2 series.]
– jpp
Nov 1 '18 at 14:43
This should be the accepted solution! [Or, if possible, just expanding the series of lists to 2 series.]
– jpp
Nov 1 '18 at 14:43
Won't this run into issues if the lists are differently sized, though perhaps that's outside of the scope of this example.
– ALollz
Nov 1 '18 at 15:40
Won't this run into issues if the lists are differently sized, though perhaps that's outside of the scope of this example.
– ALollz
Nov 1 '18 at 15:40
2
2
@ALollz yes and yes
– piRSquared
Nov 1 '18 at 15:53
@ALollz yes and yes
– piRSquared
Nov 1 '18 at 15:53
Nice! My only concern is, this solution converts datatype twice, what if my dataframe is very big, will this conversion cost more time?
– Shiang Hoo
Nov 2 '18 at 2:41
Nice! My only concern is, this solution converts datatype twice, what if my dataframe is very big, will this conversion cost more time?
– Shiang Hoo
Nov 2 '18 at 2:41
add a comment |
6
Using numpy
df.A.apply(lambda x: (np.array(x) == np.array([1,2])).all())
0 True
1 False
2 False
answered Nov 1 '18 at 14:32
VaishaliVaishali
18.6k31030
add a comment |
6
Using numpy
df.A.apply(lambda x: (np.array(x) == np.array([1,2])).all())
0 True
1 False
2 False
answered Nov 1 '18 at 14:32
VaishaliVaishali
18.6k31030
add a comment |
6
6
6
Using numpy
df.A.apply(lambda x: (np.array(x) == np.array([1,2])).all())
0 True
1 False
2 False
answered Nov 1 '18 at 14:32
VaishaliVaishali
18.6k31030
Using numpy
df.A.apply(lambda x: (np.array(x) == np.array([1,2])).all())
0 True
1 False
2 False
answered Nov 1 '18 at 14:32
VaishaliVaishali
18.6k31030
edited Nov 21 '18 at 0:28
answered Nov 1 '18 at 14:32
VaishaliVaishali
18.6k31030
answered Nov 1 '18 at 14:32
VaishaliVaishali
18.6k31030
answered Nov 1 '18 at 14:32
VaishaliVaishali
18.6k31030
18.6k31030
add a comment |
add a comment |
0
Or:
df['A'].apply(([1,2]).__eq__)
Then:
df[df['A'].apply(([1,2]).__eq__)]
answered Nov 9 '18 at 4:18
U9-ForwardU9-Forward
14k21337
add a comment |
0
Or:
df['A'].apply(([1,2]).__eq__)
Then:
df[df['A'].apply(([1,2]).__eq__)]
answered Nov 9 '18 at 4:18
U9-ForwardU9-Forward
14k21337
add a comment |
0
0
0
Or:
df['A'].apply(([1,2]).__eq__)
Then:
df[df['A'].apply(([1,2]).__eq__)]
answered Nov 9 '18 at 4:18
U9-ForwardU9-Forward
14k21337
Or:
df['A'].apply(([1,2]).__eq__)
Then:
df[df['A'].apply(([1,2]).__eq__)]
answered Nov 9 '18 at 4:18
U9-ForwardU9-Forward
14k21337
answered Nov 9 '18 at 4:18
U9-ForwardU9-Forward
14k21337
answered Nov 9 '18 at 4:18
U9-ForwardU9-Forward
14k21337
answered Nov 9 '18 at 4:18
U9-ForwardU9-Forward
14k21337
14k21337
add a comment |
add a comment |
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The only thing you're "missing" is that data frames aren't really great for storing lists. Any reason you don't want two separate columns?
– BallpointBen
Nov 1 '18 at 21:18
@BallpointBen Thanks for your attention, I've posted a new question to explain the whole question. stackoverflow.com/questions/53115592/…
– Shiang Hoo
Nov 2 '18 at 9:11
@Luuklag This may be a duplicate, but I don't believe it's a duplicate of the target you suggest. That one seems to be trying to filter based on whether multiple columns are equal to particular values. This one is trying to check if the list is equal to a single column's value, which has a very different answer.
– jpmc26
Nov 13 '18 at 22:19
Feel free to suggest a more appropriate target.
– Luuklag
Nov 13 '18 at 23:05
@Luuklag, I posted the two questions because I don't think they are the same. As jpmc described, they are connected but also very different. This post is actually the varietas of that one: I tried stupid things to solve that one and based on the stupid thing I posted this one. But this one still has its distinct value. Can you please remove the duplicate target?
– Shiang Hoo
Nov 19 '18 at 2:56