用熊猫计数和排序

Question

我有一个数据框的值形成一个文件,通过它我按两列分组,返回聚合的计数.现在我想按最大计数值排序,但是我收到以下错误:

KeyError:'count'

通过agg count列查看组是某种索引所以不知道如何做到这一点,我是Python和Panda的初学者.这是实际的代码,如果您需要更多详细信息,请与我们联系:

def answer_five():
    df = census_df#.set_index(['STNAME'])
    df = df[df['SUMLEV'] == 50]
    df = df[['STNAME','CTYNAME']].groupby(['STNAME']).agg(['count']).sort(['count'])
    #df.set_index(['count'])
    print(df.index)
    # get sorted count max item
    return df.head(5)

Answer 1

我认为你需要添加reset_index,然后参数ascending=False,sort_values因为sort返回:

FutureWarning:sort(columns = ....)已弃用,使用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)

要么:

df = df[['STNAME','CTYNAME']].groupby(['STNAME'])['CTYNAME'].size().nlargest(5)

size和之间的区别count是:

size计算NaN值,count不是.

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

Answer 2

我不知道你的df究竟是怎么样的.但是,如果您必须按其计数对几个类别的频率进行排序,则可以更轻松地从df中对系列进行切片并对系列进行排序:

series = df.count().sort_values(ascending=False)
series.head()

请注意,此系列将使用类别的名称作为索引!

Answer 3

一些现有的答案已经过时了。以下解决方案适用于列出列及其不同值的频率：

df = df[col].value_counts(ascending=False).reset_index()