Here is the sample code.
import pandas as pd, numpy as np
df = pd.DataFrame(np.random.randint(0,100,size=(10, 1)), columns=list('A'))
I have a list dl=[0,2,3,4,7]
At the index positions specified by list, I would like to have column A as "Yes".
The following code works
df.loc[dl,'A']='Yes'
How do I fill column 'A' with 'No' for column values not in index. Please forgive me if this is a duplicate post.
np.whereI'm making an assumption that there is a better way to do both 'Yes' and 'No' at the same time. If you truly just want to fill in the 'No' after you've already got the 'Yes' then refer to Fatemehhh's answer
df.loc[:, 'A'] = np.where(df.index.isin(dl), 'Yes', 'No')
Not meant for actual suggestions
f = dl.__contains__
g = ['No', 'Yes'].__getitem__
df.loc[:, 'A'] = [*map(g, map(f, df.index))]
df
A
0 Yes
1 No
2 Yes
3 Yes
4 Yes
5 No
6 No
7 Yes
8 No
9 No
One way is to use the isin function. '~' will reverse it so that the output is elements that are not in dl.
df.loc[~df.index.isin(dl),'A']='No'
First fill all rows of "A" with "No". Then update/overwrite the specific rows with "Yes".
df.loc[:,'A']='No'
df.loc[dl,'A']='Yes'
by using the difference between the list dl and the df list itself:
df.iloc[list( set(df.index) - set(dl))] = 'No'
or
df.iloc[[x for x in range(len(df)) if x not in dl]] = 'No'
pandas.Index objects already have a difference method. So this should work: df.loc[df.index.difference(dl), 'A'] = 'No'loc can handle sets just fine. To shorten up the syntax with a little set literal unpacking (not even sure what to call this) df.loc[{*df.index} - {*dl}, 'A'] = 'No'