Python 对列执行计数函数，并将字符串结果附加到列

大家好，我想计算数据集中的行数，并将总计数附加到列名 . 下面是我的数据集的外观

import pandas as pd
df = pd.DataFrame([('Jorh Hospital','2017-03-15', 389.0,34, 32, 34),
                   ('Jorh Hospital','2018-04-20', np.nan,22, 5, 43),
                   ('Jorh Hospital','2018-05-20', np.nan,22, 5, 43),
                   ('Bugh Hospital','2019-02-16', 80.5,np.nan, 56, np.nan),
                   ('Bugh Hospital','2019-03-23', np.nan,89, 67, np.nan),
                  ('Bugh Hospital','2019-04-23', np.nan,89, 67, np.nan)],
                  columns=('Hosp_name','date', 'max_rec', 'reg_pp', 'disch_no', 'temp_rec'))
df

我所尝试的是这一点，我只能在每一个单独的专栏中做。我如何一次完成所有列

df['max_rec'].count()
df['reg_pp'].count()

我不想单独这样做，我如何执行每列计数，并将结果附加到列名，就像这样的最终结果

import pandas as pd
dff = pd.DataFrame([('Jorh Hospital','2017-03-15', 389.0,34, 32, 34),
                   ('orh Hospital','2018-04-20', np.nan,22, 5, 43),
                   ('Jorh Hospital','2018-05-20', np.nan,22, 5, 43),
                   ('Bugh Hospital','2019-02-16', 80.5,np.nan, 56, np.nan),
                   ('Bugeh Hospital','2019-03-23', np.nan,89, 67, np.nan),
                  ('ugh Hospital','2019-04-23', np.nan,89, 67, np.nan)],
                  columns=('Hosp_name','date', 'max_rec N=2', 'reg_pp N=5', 'disch_no N=6', 'temp_rec N=3'))
dff

期望

dff = pd.DataFrame([('max_rec','50% (1)', '50%(1)'),
                   ('reg_pp','100%(0)', '50%(1)'),
                   ('disch_no','100%(0)', '100%(0)'),
                   ('temp_rec','100%(0)', '0')],
                  columns=('variables','Jorh Hospital (N=2)', 'Bugh Hospital (N=2)'))
dff

---

选择所有不包含前2个by and的列，然后创建包含联接值的字典并传递到重命名：

编辑：

编辑1：

df = pd.DataFrame([('Jorh Hospital','2017-03-15', 389.0,34, 32, 34),
                   ('Jorh Hospital','2018-04-20', np.nan,22, 5, 43),
                   ('Jorh Hospital','2018-05-20', np.nan,22, 5, 43),
                   ('Bugh Hospital','2019-02-16', 80.5,np.nan, 56, np.nan),
                   ('Bugh Hospital','2019-03-23', np.nan,89, 67, np.nan),
                  ('Bugh Hospital','2019-04-23', np.nan,89, 67, np.nan)],
                  columns=('Hosp_name','date', 'max_rec', 'reg_pp', 'disch_no', 'temp_rec'))

print (df)
       Hosp_name        date  max_rec  reg_pp  disch_no  temp_rec
0  Jorh Hospital  2017-03-15    389.0    34.0        32      34.0
1  Jorh Hospital  2018-04-20      NaN    22.0         5      43.0
2  Jorh Hospital  2018-05-20      NaN    22.0         5      43.0
3  Bugh Hospital  2019-02-16     80.5     NaN        56       NaN
4  Bugh Hospital  2019-03-23      NaN    89.0        67       NaN
5  Bugh Hospital  2019-04-23      NaN    89.0        67       NaN

---

A请将您的代码修改为groupby hospital_name（按医院名称分组），以便医院名称可以列为itI的附加计数I有此列，它可以按医院名称df.iloc[：，2:].notna.groupbydf['Hosp_name']对行进行分组。mean.T，您可以运行它以查看预期的数据帧，但在执行此操作之前，意味着我想先进行计数您编辑的答案正是我想要的，但您能否从索引中删除计数编号，以便将其附加到医院名称中，每家医院的计数我编辑了数据框，使其仅包括两家医院。我在最后添加了一个表格，说明了计数和执行平均值后的预期结果。所以计数是指每家医院的值，然后平均值是以百分比表示的NAN数

s = df.iloc[:, 2:].count()
d = dict(zip(s.index, s.index + ' N=' + s.astype(str)))
df = df.iloc[:,2:].notna().groupby(df['Hosp_name']).mean().T.rename(d)
print (df)
Hosp_name     Bugeh Hospital  Bugh Hospital  Jorh Hospital  orh Hospital  \
max_rec N=2              0.0            1.0            0.5           0.0   
reg_pp N=5               1.0            0.0            1.0           1.0   
disch_no N=6             1.0            1.0            1.0           1.0   
temp_rec N=3             0.0            0.0            1.0           1.0   

Hosp_name     ugh Hospital  
max_rec N=2            0.0  
reg_pp N=5             1.0  
disch_no N=6           1.0  
temp_rec N=3           0.0  

df = pd.DataFrame([('Jorh Hospital','2017-03-15', 389.0,34, 32, 34),
                   ('Jorh Hospital','2018-04-20', np.nan,22, 5, 43),
                   ('Jorh Hospital','2018-05-20', np.nan,22, 5, 43),
                   ('Bugh Hospital','2019-02-16', 80.5,np.nan, 56, np.nan),
                   ('Bugh Hospital','2019-03-23', np.nan,89, 67, np.nan),
                  ('Bugh Hospital','2019-04-23', np.nan,89, 67, np.nan)],
                  columns=('Hosp_name','date', 'max_rec', 'reg_pp', 'disch_no', 'temp_rec'))

print (df)
       Hosp_name        date  max_rec  reg_pp  disch_no  temp_rec
0  Jorh Hospital  2017-03-15    389.0    34.0        32      34.0
1  Jorh Hospital  2018-04-20      NaN    22.0         5      43.0
2  Jorh Hospital  2018-05-20      NaN    22.0         5      43.0
3  Bugh Hospital  2019-02-16     80.5     NaN        56       NaN
4  Bugh Hospital  2019-03-23      NaN    89.0        67       NaN
5  Bugh Hospital  2019-04-23      NaN    89.0        67       NaN

df = (df.iloc[:,2:]
         .notna()
         .astype(int)
         .groupby(df['Hosp_name'])
        .agg(['sum', 'mean'])
        .stack(0))
print (df)
                            mean  sum
Hosp_name                            
Bugh Hospital disch_no  1.000000    3
              max_rec   0.333333    1
              reg_pp    0.666667    2
              temp_rec  0.000000    0
Jorh Hospital disch_no  1.000000    3
              max_rec   0.333333    1
              reg_pp    1.000000    3
              temp_rec  1.000000    3

a = df['mean'].mul(100).round(0).astype(int).astype(str) + '% '
b = '(' + df['sum'].astype(str) + ')' 

s = df['sum'].sum(level=0)
d = dict(zip(s.index, s.index + ' N=' + s.astype(str)))
print (d)
{'Bugh Hospital': 'Bugh Hospital N=6', 'Jorh Hospital': 'Jorh Hospital N=10'}

df = a.add(b).unstack(0).rename(columns=d)
print (df)
Hosp_name Bugh Hospital N=6 Jorh Hospital N=10
disch_no           100% (3)           100% (3)
max_rec             33% (1)            33% (1)
reg_pp              67% (2)           100% (3)
temp_rec             0% (0)           100% (3)