如果一列与值匹配，则从 dataframe 中删除行 - Python 3.6

Question

I have a csv that looks like this:

screen_name,tweet,following,followers,is_retweet,bot
narutouz16,Grad school is lonely.,59,20,0,0
narutouz16,RT @GetMadz: Sound design in this game is 10/10 game freak lied. ,59,20,1,0
narutouz16,@hbthen3rd I know I don't.,59,20,0,0
narutouz16,"@TonyKelly95 I'm still not satisfied in the ending, even though its longer.",59,20,0,0
narutouz16,I'm currently in second place in my leaderboards in duolongo.,59,20,0,0

I am able to read this into a dataframe using the following:

df = pd.read_csv("file.csv")

That works great. I get the following dimensions when I print(df.shape) (1223726, 6)

I have a list of usernames, like below:

bad_names = ['BELOZEROVNIKIT',  'ALTMANBELINDA',    '666STEVEROGERS',   'ALVA_MC_GHEE',     'CALIFRONIAREP',    'BECCYWILL',    'BOGDANOVAO2',  'ADELE_BROCK',  'ANN1EMCCONNELL',   'ARONHOLDEN8',  'BISHOLORINE',  'BLACKTIVISTSUS',   'ANGELITHSS',   'ANWARJAMIL22',     'BREMENBOTE',   'BEN_SAR_GENT',     'ASSUNCAOWALLAS',   'AHMADRADJAB',  'AN_N_GASTON',  'BLACK_ELEVATION',  'BERT_HENLEY',  'BLACKERTHEBERR5',  'ARTHCLAUDIA',  'ALBERTA_HAYNESS',  'ADRIANAMFTTT']

What I want to do is loop through the dataframe, and if the username is in this list at all, to remove those rows from df and add them to a new df called bad_names_df .

Pseudocode would look like:

for each row in df:
    if row.username in bad_names:
        bad_names_df.append(row)
        df.remove(row)
    else:
        continue

My attempt:

for row, col in df.iterrows():
    if row['username'] in bad_user_names:
        new_df.append(row)
    else:
        continue

How is it possible to (efficiently) loop through df , with over 1.2M rows, and if the username is in the bad_names list, remove that row and add that row to a bad_names_df ? I have not found any other SO posts that address this issue.

Answer 1

You can apply a lambda then filter as follows:

df['keep'] = df['username'].apply(lambda x: False if x in bad_names else True)
df = df[df['keep']==True]

Answer 2

You can also create a mask using isin :

mask = df["screen_name"].isin(bad_names)
print (df[mask])  #df of bad names
print (df[~mask]) #df of good names

Answer 3

Using isin in one line:

bad_names_df = df[df['screen_name].isin(bad_names)]