Pandas 按一列分组并仅保留列具有集合中所有值的组

Question

I have a df as follows:

foo bar baz
aaa 0   Laos
aaa 45  Nigeria
aaa 123 Panama
bbb 12  Panama
bbb 826 Nigeria
ccc 0   Laos
ccc 15  Laos
ccc 72  Panama
ddd 4   Panama
ddd 9   Laos
ddd 987 Panama
ddd 25  Nigeria

I also have a set: {"laos", "panama", "nigeria"}

I would like to groupby("foo") and only retain the groups for which column "baz" contains all values in the set.

So, the resulting df would contain only those lines (since bbb lacks Laos and ccc lacks Nigeria):

foo bar baz
aaa 0   Laos
aaa 45  Nigeria
aaa 123 Panama
ddd 4   Panama
ddd 9   Laos
ddd 987 Panama
ddd 25  Nigeria

Answer 1

Try with

s=df.groupby('foo').\
      filter(lambda x : pd.Series(["laos", "panama", "nigeria"]).isin(x['baz'].str.lower()).all())
Out[21]: 
    foo  bar      baz
0   aaa    0     Laos
1   aaa   45  Nigeria
2   aaa  123   Panama
8   ddd    4   Panama
9   ddd    9     Laos
10  ddd  987   Panama
11  ddd   25  Nigeria

Answer 2

IIUC, Series.str.lower with Series.isin and GroupBy.transform

l = ["laos", "panama", "nigeria"]
s = df['baz'].str.lower()

m = (s.isin(l)
      .mask(df.duplicated(['baz', 'foo']), False)
      .groupby(df['foo'])
      .transform('sum').eq(len(l)))

df_filtered = df.loc[m]
print(df_filtered)


    foo  bar      baz
0   aaa    0     Laos
1   aaa   45  Nigeria
2   aaa  123   Panama
8   ddd    4   Panama
9   ddd    9     Laos
10  ddd  987   Panama
11  ddd   25  Nigeria

It is similar to:

m = ((s.isin(l) & (~df.duplicated(['baz', 'foo'])))
       .groupby(df['foo'])
       .transform('sum').eq(len(l)))

Answer 3

df1 = df[df.groupby('foo')['baz'].transform('nunique').eq(3)]