使用df.loc []和In

Question

I am trying to do the following code :

df.loc[df['Column'] in list]=1
df.loc[df['Column'] not in list]=2

But I have the famous error

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

So from what I understood from this error I already got, it is linked with the logic. What I want to know is Is there a specific typo to do what I am trying to do with a list, or do I have to expand my entire list like that (which in my case would be pretty ugly and long) :

df.loc[df[('Column']=='a') & (df['Column']=='b')]=1
df.loc[df[('Column']!='a') & (df['Column']!='b')]=2 

Answer 1

I'd use np.where . However, you need a column name.

df['new_column'] = np.where(df['Column'].isin(lst), 1, 2)

Consider the list lst and dataframe df

lst = [1, 2, 3]
df = pd.DataFrame(dict(Column=[0, 1, 2, 3, 4]))

Then

df['new_column'] = np.where(df['Column'].isin(lst), 1, 2)

print(df)

   Column  new_column
0       0           2
1       1           1
2       2           1
3       3           1
4       4           2

Answer 2

您正在寻找的功能是

df.loc[df['Column'].isin(list)] = 1