I want to creat a column s['C']
using apply() with a Pandas DataFrame.
My dataset is similiar to this:
[In]:
s=pd.DataFrame({'A':['hello', 'good', 'my', 'pandas','wrong'],
'B':[['all', 'say', 'hello'],
['good', 'for', 'you'],
['so','hard'],
['pandas'],
[]]})
[Out]:
A B
0 hello [all, say, hello]
1 good [good, for, you]
2 my [so, hard]
3 pandas [pandas]
4 wrong []
I need to creat a s['C'] column where the value of each row is a list with ones and zeros dependending if the word of column A is in the list of column B and the position of the element in the list of column B. My output should be like this:
[Out]:
A B C
0 hello [all, say, hello] [0, 0, 1]
1 good [good, for, you] [1, 0, 0]
2 my [so, hard] [0, 0]
3 pandas [pandas] [1]
4 wrong [] [0]
I've been trying with a función and apply, but I still have not realized where is the error.
[In]:
def func(valueA,listB):
new_list=[]
for i in listB:
if listB[i] == valueA:
new_list.append(1)
else:
new_list.append(0)
return new_list
s['C']=s.apply( lambda x: func(x.loc[:,'A'], x.loc[:,'B']))
The error is: Too many indexers
And I also tried with:
[In]:
list=[]
listC=[]
for i in s['A']:
for j in s['B'][i]:
if s['A'][i] == s['B'][i][j]:
list.append(1)
else:
list.append(0)
listC.append(list)
s['C']=listC
The error is: KeyError: 'hello'
Any suggests?
If you are working with pandas 0.25+, explode
is an option:
(s.explode('B')
.assign(C=lambda x: x['A'].eq(x['B']).astype(int))
.groupby(level=0).agg({'A':'first','B':list,'C':list})
)
Output:
A B C
0 hello [all, say, hello] [0, 0, 1]
1 good [good, for, you] [1, 0, 0]
2 my [so, hard] [0, 0]
3 pandas [pandas] [1]
4 wrong [nan] [0]
Option 2: Based on your logic, you can do a list comprehension. This should work with any version of pandas
:
s['C'] = [[x==a for x in b] if b else [0] for a,b in zip(s['A'],s['B'])]
Output:
A B C
0 hello [all, say, hello] [False, False, True]
1 good [good, for, you] [True, False, False]
2 my [so, hard] [False, False]
3 pandas [pandas] [True]
4 wrong [] [0]
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