I have a dataframe containing column of lists:
col_1
[A, A, A, B, C]
[D, B, C]
[C]
[A, A, A]
NaN
I want to create new column, if the list starts with 3* A return 1, if not return 0:
col_1 new_col
[A, A, A, B, C] 1
[D, B, C] 0
[C] 0
[A, A, A] 1
NaN 0
I tried this but didn't work:
df['new_col'] = df.loc[df.col_1[0:3] == [A, A, A]]
Beause there are some non list values is possible use if-else lambda function for 0 if not list:
print (df['col_1'].map(type))
0 <class 'list'>
1 <class 'list'>
2 <class 'list'>
3 <class 'list'>
4 <class 'float'>
Name: col_1, dtype: object
f = lambda x: int((x[:3]) == ['A','A','A']) if isinstance(x, list) else 0
df['new_col'] = df['col_1'].map(f)
#alternative
#df['new_col'] = df['col_1'].apply(f)
print (df)
col_1 new_col
0 [A, A, A, B, C] 1
1 [D, B, C] 0
2 [C] 0
3 [A, A, A] 1
4 NaN 0
print (df['col_1'].map(type)) ?df['col_1'] = df['col_1'].str.strip('[]').str.split(',') before my solution for listsHere is another potential solution using map:
import pandas as pd
#borrowing dataframe from @Alexendra
df = pd.DataFrame({
'col_1': [
['A', 'A', 'A', 'B', 'C'],
['D', 'B', 'C'],
['C'],
['A', 'A', 'A']
]
})
df['new_col'] = df['col_1'].map(lambda x : 1 if x[0:3] == ['A','A','A'] else 0)
print(df)
Output:
col_1 new_col
0 [A, A, A, B, C] 1
1 [D, B, C] 0
2 [C] 0
3 [A, A, A] 1
Solution via apply lambda:
df = pd.DataFrame({
'col_1': [
['A', 'A', 'A', 'B', 'C'],
['D', 'B', 'C'],
['C'],
['A', 'A', 'A']
]
})
df['new_col'] = df.col_1.apply(lambda x: x[0:3] == ['A', 'A', 'A'] if isinstance(x, list) else False).view('i1')
df.head()
Output:
