I have a DataFrame like
name col1 col2
a aa 123
a bb 123
b aa 234
and a list
[1, 2, 3]
I want to replace the col2 of every row with col1 = 'aa' with the list like
name col1 col2
a aa [1, 2, 3]
a bb 123
b aa [1, 2, 3]
I tried something like
df.loc[df[col1] == 'aa', col2] = [1, 2, 3]
but it gives me the error:
ValueError: could not broadcast input array from shape (xx,) into shape (yy,)
How should I get around this?
A list comprehension with an if/else should work
df['col2'] = [x['col2'] if x['col1'] != 'aa' else [1,2,3] for ind,x in df.iterrows()]
It will be safe to do with for loop
df.col2 = df.col2.astype(object)
for x in df.index:
if df.at[x,'col1'] == 'aa':
df.at[x,'col2'] = [1,2,3]
df
name col1 col2
0 a aa [1, 2, 3]
1 a bb 123
2 b aa [1, 2, 3]
You can also use:
data = {'aa':[1,2,3]}
df['col2'] = np.where(df['col1'] == 'aa', df['col1'].map(data), df['col2'])
You should use this with care, as doing this will change list to both locations:
df['col2'].loc[0].append(5)
print(df)
#OUTPUT
name col1 col2
0 a aa [1, 2, 3, 5]
1 a bb 123
2 b aa [1, 2, 3, 5]
But this is fine:
df = df.loc[1:]
print(df)
#OUTPUT
name col1 col2
1 a bb 123
2 b aa [1, 2, 3]
Make it simple. np.where should do. Code below
df['col2']=np.where(df['col1']=='aa', str(lst), df['col2'])
Alternatively use pd.Series with list locked in double brackects
df['col2']=np.where(df['col1']=='aa', pd.Series([lst]), df['col2'])
import pandas as pd
df = pd.DataFrame({"name":["a","a","b"],"col1":["aa","bb","aa"],"col2":[123,123,234]})
l = [1,2,3]
df["col2"] = df.apply(lambda x: l if x.col1 == "aa" else x.col2, axis =1)
df