我想你需要添加reset_index
,然后参数ascending=False
到sort_values
因为sort
回报:
FutureWarning: sort(columns=....) is deprecated, use sort_values(by=.....) .sort_values(['count'], ascending=False)
df = df[['STNAME','CTYNAME']].groupby(['STNAME'])['CTYNAME'] \
.count() \
.reset_index(name='count') \
.sort_values(['count'], ascending=False) \
.head(5)
样品:
df = pd.DataFrame({'STNAME':list('abscscbcdbcsscae'),
'CTYNAME':[4,5,6,5,6,2,3,4,5,6,4,5,4,3,6,5]})
print (df)
CTYNAME STNAME
0 4 a
1 5 b
2 6 s
3 5 c
4 6 s
5 2 c
6 3 b
7 4 c
8 5 d
9 6 b
10 4 c
11 5 s
12 4 s
13 3 c
14 6 a
15 5 e
df = df[['STNAME','CTYNAME']].groupby(['STNAME'])['CTYNAME'] \
.count() \
.reset_index(name='count') \
.sort_values(['count'], ascending=False) \
.head(5)
print (df)
STNAME count
2 c 5
5 s 4
1 b 3
0 a 2
3 d 1
但似乎你需要Series.nlargest
:
df = df[['STNAME','CTYNAME']].groupby(['STNAME'])['CTYNAME'].count().nlargest(5)
或:
df = df[['STNAME','CTYNAME']].groupby(['STNAME'])['CTYNAME'].size().nlargest(5)
The difference between size
and count
is:
size
counts NaN
values, count
does not.
样品:
df = pd.DataFrame({'STNAME':list('abscscbcdbcsscae'),
'CTYNAME':[4,5,6,5,6,2,3,4,5,6,4,5,4,3,6,5]})
print (df)
CTYNAME STNAME
0 4 a
1 5 b
2 6 s
3 5 c
4 6 s
5 2 c
6 3 b
7 4 c
8 5 d
9 6 b
10 4 c
11 5 s
12 4 s
13 3 c
14 6 a
15 5 e
df = df[['STNAME','CTYNAME']].groupby(['STNAME'])['CTYNAME']
.size()
.nlargest(5)
.reset_index(name='top5')
print (df)
STNAME top5
0 c 5
1 s 4
2 b 3
3 a 2
4 d 1
很好,谢谢你解释各种选项 – Rubans