Pandas, isin, column of lists
Trying to make a Boolean flag that reads TRUE if one value or another is within a list. The below code is returning a FALSE for row 1 and I am not sure why, could someone help me understand why a FALSE is getting returned for the first row?
lists={'someList!':[[1,2,12,6,'ABC'],[1000,4,'z','a','bob']]}
dfLists = pd.DataFrame(lists)
dfLists['contains?']=dfLists['someList!'].isin([0,1])
pandas
asked Nov 13 '18 at 16:43
reality stiltsreality stilts
435
add a comment |
Trying to make a Boolean flag that reads TRUE if one value or another is within a list. The below code is returning a FALSE for row 1 and I am not sure why, could someone help me understand why a FALSE is getting returned for the first row?
lists={'someList!':[[1,2,12,6,'ABC'],[1000,4,'z','a','bob']]}
dfLists = pd.DataFrame(lists)
dfLists['contains?']=dfLists['someList!'].isin([0,1])
pandas
asked Nov 13 '18 at 16:43
reality stiltsreality stilts
435
Your code gives an error instead. Please provide a reproducible code.
– TeeKea
Nov 13 '18 at 16:54
add a comment |
Trying to make a Boolean flag that reads TRUE if one value or another is within a list. The below code is returning a FALSE for row 1 and I am not sure why, could someone help me understand why a FALSE is getting returned for the first row?
lists={'someList!':[[1,2,12,6,'ABC'],[1000,4,'z','a','bob']]}
dfLists = pd.DataFrame(lists)
dfLists['contains?']=dfLists['someList!'].isin([0,1])
pandas
asked Nov 13 '18 at 16:43
reality stiltsreality stilts
435
Trying to make a Boolean flag that reads TRUE if one value or another is within a list. The below code is returning a FALSE for row 1 and I am not sure why, could someone help me understand why a FALSE is getting returned for the first row?
lists={'someList!':[[1,2,12,6,'ABC'],[1000,4,'z','a','bob']]}
dfLists = pd.DataFrame(lists)
dfLists['contains?']=dfLists['someList!'].isin([0,1])
pandas
pandas
asked Nov 13 '18 at 16:43
reality stiltsreality stilts
435
asked Nov 13 '18 at 16:43
reality stiltsreality stilts
435
asked Nov 13 '18 at 16:43
reality stiltsreality stilts
435
asked Nov 13 '18 at 16:43
reality stiltsreality stilts
435
asked Nov 13 '18 at 16:43
reality stiltsreality stilts
435
435
Your code gives an error instead. Please provide a reproducible code.
– TeeKea
Nov 13 '18 at 16:54
add a comment |
Your code gives an error instead. Please provide a reproducible code.
– TeeKea
Nov 13 '18 at 16:54
Your code gives an error instead. Please provide a reproducible code.
– TeeKea
Nov 13 '18 at 16:54
Your code gives an error instead. Please provide a reproducible code.
– TeeKea
Nov 13 '18 at 16:54
add a comment |
3 Answers
3
active
oldest
votes
Could someone help me understand why a FALSE is getting returned for the first row?
This isn't working because .isin(values) returns whether each element in the Series is contained in values.
You can use {0, 1} as a set and apply the truthiness of its intersection to each list:
>>> s = {0, 1}
>>> dfLists['someList!'].apply(lambda x: bool(s.intersection(x)))
0 True
1 False
This effectively does:
>>> s.intersection([1, 2, 12, 6, 'ABC'])
{1}
>>> s.intersection([1000, 4, 'z', 'a', 'bob'])
set()
The bool of the first result is True, because it is non-empty.
answered Nov 13 '18 at 16:54
Brad SolomonBrad Solomon
13.7k73484
add a comment |
Using Dataframe constructor flatten you list column, then using isin
pd.DataFrame(dfLists['someList!'].tolist()).isin([1,2]).any(1)
Out[39]:
0 True
1 False
dtype: bool
answered Nov 13 '18 at 16:55
Wen-BenWen-Ben
1
thx for this answer, found it helpful
– reality stilts
Nov 13 '18 at 17:58
add a comment |
You are passing the dataframe a list of lists. It's comparing integers to lists, so it's not finding a match.
Define your columns distinctly, like this, and isin should work.
lists={'someList!':[1,2,12,6,'ABC'], 'someList2':[1000,4,'z','a','bob']}
answered Nov 13 '18 at 16:56
CobraCobra
225
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Could someone help me understand why a FALSE is getting returned for the first row?
This isn't working because .isin(values) returns whether each element in the Series is contained in values.
You can use {0, 1} as a set and apply the truthiness of its intersection to each list:
>>> s = {0, 1}
>>> dfLists['someList!'].apply(lambda x: bool(s.intersection(x)))
0 True
1 False
This effectively does:
>>> s.intersection([1, 2, 12, 6, 'ABC'])
{1}
>>> s.intersection([1000, 4, 'z', 'a', 'bob'])
set()
The bool of the first result is True, because it is non-empty.
answered Nov 13 '18 at 16:54
Brad SolomonBrad Solomon
13.7k73484
add a comment |
Could someone help me understand why a FALSE is getting returned for the first row?
This isn't working because .isin(values) returns whether each element in the Series is contained in values.
You can use {0, 1} as a set and apply the truthiness of its intersection to each list:
>>> s = {0, 1}
>>> dfLists['someList!'].apply(lambda x: bool(s.intersection(x)))
0 True
1 False
This effectively does:
>>> s.intersection([1, 2, 12, 6, 'ABC'])
{1}
>>> s.intersection([1000, 4, 'z', 'a', 'bob'])
set()
The bool of the first result is True, because it is non-empty.
answered Nov 13 '18 at 16:54
Brad SolomonBrad Solomon
13.7k73484
add a comment |
Could someone help me understand why a FALSE is getting returned for the first row?
This isn't working because .isin(values) returns whether each element in the Series is contained in values.
You can use {0, 1} as a set and apply the truthiness of its intersection to each list:
>>> s = {0, 1}
>>> dfLists['someList!'].apply(lambda x: bool(s.intersection(x)))
0 True
1 False
This effectively does:
>>> s.intersection([1, 2, 12, 6, 'ABC'])
{1}
>>> s.intersection([1000, 4, 'z', 'a', 'bob'])
set()
The bool of the first result is True, because it is non-empty.
answered Nov 13 '18 at 16:54
Brad SolomonBrad Solomon
13.7k73484
Could someone help me understand why a FALSE is getting returned for the first row?
This isn't working because .isin(values) returns whether each element in the Series is contained in values.
You can use {0, 1} as a set and apply the truthiness of its intersection to each list:
>>> s = {0, 1}
>>> dfLists['someList!'].apply(lambda x: bool(s.intersection(x)))
0 True
1 False
This effectively does:
>>> s.intersection([1, 2, 12, 6, 'ABC'])
{1}
>>> s.intersection([1000, 4, 'z', 'a', 'bob'])
set()
The bool of the first result is True, because it is non-empty.
answered Nov 13 '18 at 16:54
Brad SolomonBrad Solomon
13.7k73484
answered Nov 13 '18 at 16:54
Brad SolomonBrad Solomon
13.7k73484
answered Nov 13 '18 at 16:54
Brad SolomonBrad Solomon
13.7k73484
answered Nov 13 '18 at 16:54
Brad SolomonBrad Solomon
13.7k73484
13.7k73484
add a comment |
add a comment |
Using Dataframe constructor flatten you list column, then using isin
pd.DataFrame(dfLists['someList!'].tolist()).isin([1,2]).any(1)
Out[39]:
0 True
1 False
dtype: bool
answered Nov 13 '18 at 16:55
Wen-BenWen-Ben
1
thx for this answer, found it helpful
– reality stilts
Nov 13 '18 at 17:58
add a comment |
Using Dataframe constructor flatten you list column, then using isin
pd.DataFrame(dfLists['someList!'].tolist()).isin([1,2]).any(1)
Out[39]:
0 True
1 False
dtype: bool
answered Nov 13 '18 at 16:55
Wen-BenWen-Ben
1
thx for this answer, found it helpful
– reality stilts
Nov 13 '18 at 17:58
add a comment |
Using Dataframe constructor flatten you list column, then using isin
pd.DataFrame(dfLists['someList!'].tolist()).isin([1,2]).any(1)
Out[39]:
0 True
1 False
dtype: bool
answered Nov 13 '18 at 16:55
Wen-BenWen-Ben
1
Using Dataframe constructor flatten you list column, then using isin
pd.DataFrame(dfLists['someList!'].tolist()).isin([1,2]).any(1)
Out[39]:
0 True
1 False
dtype: bool
answered Nov 13 '18 at 16:55
Wen-BenWen-Ben
1
answered Nov 13 '18 at 16:55
Wen-BenWen-Ben
1
answered Nov 13 '18 at 16:55
Wen-BenWen-Ben
1
answered Nov 13 '18 at 16:55
Wen-BenWen-Ben
1
1
thx for this answer, found it helpful
– reality stilts
Nov 13 '18 at 17:58
add a comment |
thx for this answer, found it helpful
– reality stilts
Nov 13 '18 at 17:58
thx for this answer, found it helpful
– reality stilts
Nov 13 '18 at 17:58
thx for this answer, found it helpful
– reality stilts
Nov 13 '18 at 17:58
add a comment |
You are passing the dataframe a list of lists. It's comparing integers to lists, so it's not finding a match.
Define your columns distinctly, like this, and isin should work.
lists={'someList!':[1,2,12,6,'ABC'], 'someList2':[1000,4,'z','a','bob']}
answered Nov 13 '18 at 16:56
CobraCobra
225
add a comment |
You are passing the dataframe a list of lists. It's comparing integers to lists, so it's not finding a match.
Define your columns distinctly, like this, and isin should work.
lists={'someList!':[1,2,12,6,'ABC'], 'someList2':[1000,4,'z','a','bob']}
answered Nov 13 '18 at 16:56
CobraCobra
225
add a comment |
You are passing the dataframe a list of lists. It's comparing integers to lists, so it's not finding a match.
Define your columns distinctly, like this, and isin should work.
lists={'someList!':[1,2,12,6,'ABC'], 'someList2':[1000,4,'z','a','bob']}
answered Nov 13 '18 at 16:56
CobraCobra
225
You are passing the dataframe a list of lists. It's comparing integers to lists, so it's not finding a match.
Define your columns distinctly, like this, and isin should work.
lists={'someList!':[1,2,12,6,'ABC'], 'someList2':[1000,4,'z','a','bob']}
answered Nov 13 '18 at 16:56
CobraCobra
225
answered Nov 13 '18 at 16:56
CobraCobra
225
answered Nov 13 '18 at 16:56
CobraCobra
225
answered Nov 13 '18 at 16:56
CobraCobra
225
225
add a comment |
add a comment |
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Your code gives an error instead. Please provide a reproducible code.
– TeeKea
Nov 13 '18 at 16:54