Сначала создайте логическую маску, сравнив оба столбца с &
для побитового значения AND
, а затем преобразуйте в числовое значение с совокупностью sum
:
s = df['Planted by Govt'].eq('Yes') & df['Planted by College'].eq('No')
out = s.view('i1').groupby(df['Tree Name']).sum()
#alternative
#out = s.astype(int).groupby(df['Tree Name']).sum()
print (out)
Tree Name
A 1
B 3
C 2
dtype: int8
Последнее для пользовательского вывода: f-string
s:
for k, v in out.items():
print (f"{v} Tree(s) {k} were planted by govt and not by college")
1 Tree(s) A were planted by govt and not by college
3 Tree(s) B were planted by govt and not by college
2 Tree(s) C were planted by govt and not by college
Другая идея - создать новый столбец для оригинала:
df['new'] = (df['Planted by Govt'].eq('Yes') & df['Planted by College'].eq('No')).view('i1')
print (df)
Tree Name Planted by Govt Planted by College new
0 A Yes No 1
1 B Yes No 1
2 C Yes No 1
3 C Yes No 1
4 A No No 0
5 B No Yes 0
6 B Yes Yes 0
7 B Yes No 1
8 B Yes No 1
out = df.groupby('Tree Name')['new'].sum()
print (out)
Tree Name
A 1
B 3
C 2
Name: new, dtype: int8