How to set to the column list with the number of elements equal to the number of elements in the list on another column Here is the df
t1 t2
[1,2] NaN
[1] NaN
[1,2,3]NaN
I want to get
t1 t2
[1,2] [0,0]
[1] [0]
[1,2,3][0,0,0]
Here is my code
df_1['t2'][df_1['t2'].isnull() & df_1['t1'].notnull()] = [0 for i in df_1['t1']]
But it somehow doesn't return lists, only integer number
Assign mask to variable for each side of assignement and then use apply for replace values of lists (with mask for replace NaN if non NaN in t1):
import ast
#if possible strings repr of lists
#df_1['t1'] = df_1['t1'].apply(ast.literal_eval)
mask = df_1['t2'].isnull() & df_1['t1'].notnull()
df_1.loc[mask, 't2'] = df_1.loc[mask, 't1'].apply(lambda x: [0 for _ in x])
print (df_1)
t1 t2
0 [1, 2] [0, 0]
1 [1] [0]
2 [1, 2, 3] [0, 0, 0]
Alternative with np.repeat:
mask = df_1['t2'].isnull() & df_1['t1'].notnull()
df_1.loc[mask, 't2'] = df_1.loc[mask, 't1'].apply(lambda x: np.repeat(0, len(x)))
print (df_1)
t1 t2
0 [1, 2] [0, 0]
1 [1] [0]
2 [1, 2, 3] [0, 0, 0]
You can do it like this:
df['t2'] = df['t1'].apply(lambda x: [0]*len(x))
t1 t2
0 [1, 2] [0, 0]
1 [1] [0]
2 [1, 2, 3] [0, 0, 0]