If value in dataframe row = list (not in order), change another row value to 'ONLINE'

Question

import numpy as np
import pandas as pd

list1=pd.DataFrame({'A':['a','b','c','d','x'],'B':['a','b','c','d','x']})
list2=[['1','b','c','a']]
list3=['a','b','c','d','e']
list1

    A   B
0   a   a
1   b   b
2   c   c
3   d   d
4   x   x

i get a value error:

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

When I do replace list1.B using list3: it works.

for i in range(len(list1.A)):
    if list1.A[i] in list3:
        list1.B[i]='ONLINE'
list1

    A   B
0   a   ONLINE
1   b   ONLINE
2   c   ONLINE
3   d   ONLINE
4   x   x

How do I do for list2? Thanks I tried query and isin too doesn't work.

jezrael · Accepted Answer · 2020-04-12T11:42:52.330

Use:

#change DataFrame to df for distingush between lists and df
df=pd.DataFrame({'A':['a','b','c','d','x'],'B':['a','b','c','d','x']})
list2=[['1','b','c','a']]
list3=['a','b','c','d','e']

I think here is better use Series.isin with DataFrame.loc for select column by mask:

df.loc[df['B'].isin(list3), 'B'] = 'ONLINE'

For second list is nested, so select first one:

print (list2[0])
['1', 'b', 'c', 'a']
df.loc[df['B'].isin(list2[0]), 'B'] = 'ONLINE'
print (df)
   A       B
0  a  ONLINE
1  b  ONLINE
2  c  ONLINE
3  d       d
4  x       x

If there is multiple nested lists use flattening lists:

list4=[['1','b'],['c','a']]

print ([y for x in list4 for y in x])
df.loc[df['B'].isin([y for x in list4 for y in x]), 'B'] = 'ONLINE'
print (df)
   A       B
0  a  ONLINE
1  b  ONLINE
2  c  ONLINE
3  d       d
4  x       x

thanks! if there's 40000 rows, is there a better way for this? — Ricky, Apr 12 '20 at 13:57
@Ricky - I think for 40k rows all solution from my answer working nice and fast. — jezrael, Apr 13 '20 at 06:50

If value in dataframe row = list (not in order), change another row value to 'ONLINE'

1 Answers1