Pandas在字典中计算值的出现次数

Question

给定一个数据框：

pd.DataFrame({"A":[1,2,3],
              "B": [{"Mon":"Closed", "Tue":"Open", "Wed":"Closed"},
                    {"Mon":"Open", "Tue":"Open", "Wed":"Closed"},
                    {"Mon":"Open", "Tue":"Open", "Wed":"Open"}]
              })

如何统计字典中"Closed"出现的次数？

A  B    count
1 {..}  2
2 {..}  1 
3 {..}  0

我真的不知道该如何入手尝试这个。

- jxn

Answer 1

您可以使用apply：

df['count'] = df.B.apply(pd.Series).eq('Closed').sum(1)

输出：

   A                                                  B  count
0  1  {'Mon': 'Closed', 'Tue': 'Open', 'Wed': 'Closed'}      2
1  2    {'Mon': 'Open', 'Tue': 'Open', 'Wed': 'Closed'}      1
2  3      {'Mon': 'Open', 'Tue': 'Open', 'Wed': 'Open'}      0

- Quang Hoang

Answer 2

您可以尝试将字典系列转换为数据框，然后使用stack方法，接着在level=0上对Closed的值进行求和以获得每行的计数：

df['Count_closed'] = pd.DataFrame(df['B'].tolist()).stack().eq("Closed").sum(level=0)

---

   A                                                  B  Count_closed
0  1  {'Mon': 'Closed', 'Tue': 'Open', 'Wed': 'Closed'}           2.0
1  2    {'Mon': 'Open', 'Tue': 'Open', 'Wed': 'Closed'}           1.0
2  3      {'Mon': 'Open', 'Tue': 'Open', 'Wed': 'Open'}           0.0

- anky

Answer 3

我会做

df.B.astype(str).str.count('Closed')
Out[30]: 
0    2
1    1
2    0
Name: B, dtype: int64

或者

df['Cnt']=pd.DataFrame(df.B.tolist()).eq('Closed').sum(1).values
Out[35]: 
0    2
1    1
2    0
dtype: int64

- BENY

---

@anky 哈哈，我有时候也会忘记带轴的总和~ - BENY

Answer 4

直接的 .apply() 解决方案:

df['Count'] = df.B.apply(lambda x: sum('Closed' in v for v in x.values()))
print(df)

输出：

   A                                                  B  Count
0  1  {'Mon': 'Closed', 'Tue': 'Open', 'Wed': 'Closed'}      2
1  2    {'Mon': 'Open', 'Tue': 'Open', 'Wed': 'Closed'}      1
2  3      {'Mon': 'Open', 'Tue': 'Open', 'Wed': 'Open'}      0

---

基准测试：

import perfplot
import pandas as pd


def f1(df):
    df['Count'] = df.B.apply(lambda x: sum('Closed' in v for v in x.values()))
    return df

def f2(df):
    df['count'] = df.B.astype(str).str.count('Closed')
    return df

# Commented out because of timed-out:
# def f3(df):
#     df['count'] = df.B.apply(pd.Series).eq('Closed').sum(1)
#     return df

def f4(df):
    df['count'] = pd.DataFrame(df['B'].tolist()).stack().eq("Closed").sum(level=0)
    return df

def setup(n):
    A = [*range(n)]
    B = [{'Mon': 'Closed', 'Tue': 'Open', 'Wed': 'Closed'} for _ in range(n)]
    df = pd.DataFrame({'A': A,
                       'B': B})
    return df

perfplot.show(
    setup=setup,
    kernels=[f1, f2, f4],
    labels=['apply(sum)', 'str.count()', 'stack.eq()'],
    n_range=[10**i for i in range(1, 7)],
    xlabel='N (* len(df))',
    equality_check=None,
    logx=True,
    logy=True)

结果：

所以看起来直接使用 apply() 和 sum() 是最快的。

- Andrej Kesely

Answer 5

请不要将字典放入数据框列中。这样会失去向量化操作的速度，并使值难以访问。

清理您的 df:

>>> df = pd.concat([df['A'], df['B'].apply(pd.Series)], axis=1)
>>> df 
   A     Mon   Tue     Wed
0  1  Closed  Open  Closed
1  2    Open  Open  Closed
2  3    Open  Open    Open

现在统计 'Closed' 状态非常容易。

>>> df['count'] = df.eq('Closed').sum(1)
>>> df
   A     Mon   Tue     Wed  count
0  1  Closed  Open  Closed      2
1  2    Open  Open  Closed      1
2  3    Open  Open    Open      0

- timgeb

Answer 6

使用一个辅助函数：

def aux_func(x):

    week_days = x.keys()
    count=0
    for day in week_days:
        if x[day]=='Closed':
            count+=1

    return count

counts = [aux_func(c) for c in df.loc[:,'B'] ]

df['counts'] = counts

- mangelfdz

Answer 7

你可以在简单的列表推导式中使用计数器。

from collections import Counter

df['count'] = [Counter(x.values())['Closed'] for x in df.B]

#   A                                                  B  Count
#0  1  {'Mon': 'Closed', 'Tue': 'Open', 'Wed': 'Closed'}      2
#1  2    {'Mon': 'Open', 'Tue': 'Open', 'Wed': 'Closed'}      1
#2  3      {'Mon': 'Open', 'Tue': 'Open', 'Wed': 'Open'}      0

- ALollz