用 pandas 计数和排序 [英] Count and Sort with Pandas

问题描述

我有一个数据框，用于将值形成一个文件，通过该文件，我已按两列进行分组，这些列返回汇总的计数.现在，我想按最大计数值进行排序，但是出现以下错误:

I have a dataframe for values form a file by which I have grouped by two columns, which return a count of the aggregation. Now I want to sort by the max count value, however I get the following error:

KeyError:计数"

KeyError: 'count'

通过agg count列查看group是某种索引，因此不确定如何执行此操作，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，请告诉我:

Looks the group by agg count column is some sort of index so not sure how to do this, I'm a beginner to Python and Panda. Here's the actual code, please let me know if you need more detail:

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)

推荐答案

我认为您需要添加reset_index，然后将参数ascending=False添加到

I think you need add reset_index, then parameter ascending=False to sort_values because sort return:

FutureWarning:不建议使用sort(columns = ....)，请使用sort_values(by = .....) .sort_values(['count']，ascending = False)

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)

示例:

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

---

但是似乎您需要 Series.nlargest :

---

But it seems you need Series.nlargest:

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

或:

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

size和count之间的区别是:

The difference between size and count is:

size 计算NaN值，

size counts NaN values, count does not.

示例:

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

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