Pandas Python how to delete rows based on two column conditions?

Here's a short solution:

new_df = df.groupby('MultiID').filter(lambda g: ~g['Status'].eq('B').any()) 

Output:

>>> new_df Status MultiID 2 A 2 3 C 2 4 C 2 

Here's one way without groupby. Get the "MultiID"s of rows where Status=="B", then filter the rows without those MultiIDs:

MultiIDs_w_Bs = df.loc[df['Status'].eq('B'), 'MultiID'] out = df[~df['MultiID'].isin(MultiIDs_w_Bs)] 

Output:

 Status MultiID 2 A 2 3 C 2 4 C 2 

No need groupby

out = df.loc[~df.MultiID.isin(df.loc[df.Status.eq('B'),'MultiID'])] Out[169]: Status MultiID 2 A 2 3 C 2 4 C 2 

Use groupby before filter out your dataframe according your condition. Here transform allows to broadcast boolean result to each member of the group.

out = df[df.groupby('MultiID')['Status'].transform(lambda x: all(x != 'B'))] print(out) # Output Status MultiID 2 A 2 3 C 2 4 C 2 

Try:

df[~df['Status'].eq('B').groupby(df['MultiID']).transform('any')] 

Output:

 Status MultiID 2 A 2 3 C 2 4 C 2 

Details:

• Create a boolean series where status equals to B.
• Group that series by MultiID
• Using transform if any record in that group is True, the make the whole group True
• Invert True and False and boolean filter the original dataframe

s = ['A', 'B', 'A', 'C', 'C', 'A', 'B', 'A'] m = [1, 1, 2, 2, 2, 3, 3, 3] # Loop through lists and remove the element if its status is B for i in range(len(m)-1): if s[i] == 'B': del m[i] # Remove all B's from s while 'B' in s: s.remove('B') print(s) print(m) 

gives

['A', 'A', 'C', 'C', 'A', 'A'] [1, 2, 2, 2, 3, 3]