取消堆栈并返回每个变量的值计数?

Question

我有一个数据框，通过多项选择题记录了19717人选择编程语言的回答。第一列当然是受访者的性别，其余的是他们选择的选择。数据框如下所示，每个响应都记录为与列相同的名称。如果未选择响应，则会导致NaN。

ID     Gender              Python    Bash    R    JavaScript    C++
0      Male                Python    nan     nan  JavaScript    nan
1      Female              nan       nan     R    JavaScript    C++
2      Prefer not to say   Python    Bash    nan  nan           nan
3      Male                nan       nan     nan  nan           nan

我想要的是一个基于Gender返回计数的表。因此，如果有 5000 名男性用 Python 编码，3000 名女性用 JS，那么我应该得到这个：

Gender              Python    Bash    R    JavaScript    C++
Male                5000      1000    800  1500          1000
Female              4000      500     1500 3000          800
Prefer Not To Say   2000      ...   ...    ...           860

我尝试了一些选项：

df.iloc[:, [*range(0, 13)]].stack().value_counts()
Male                       16138
Python                     12841
SQL                         6532
R                           4588
Female                      3212
Java                        2267
C++                         2256
Javascript                  2174
Bash                        2037
C                           1672
MATLAB                      1516
Other                       1148
TypeScript                   389
Prefer not to say            318
None                          83
Prefer to self-describe       49
dtype: int64

这不是如上所述的要求。这可以在大熊猫身上完成吗？

Answer 1

另一个想法是沿轴 1applyjoin值，get_dummies然后groupby：

(df.loc[:, 'Python':]
.apply(lambda x: '|'.join(x.dropna()), axis=1)
.str.get_dummies('|')
.groupby(df['Gender']).sum())

[出]

Bash  C++  JavaScript  Python  R
Gender                                             
Female                0    1           1       0  1
Male                  0    0           1       1  0
Prefer not to say     1    0           0       1  0

Answer 2

您可以将Gender设置为索引和总和：

s = df.set_index('Gender').iloc[:, 1:]
s.eq(s.columns).astype(int).sum(level=0)

输出：

Python  Bash  R  JavaScript  C++
Gender                                             
Male                    1     0  0           1    0
Female                  0     0  1           1    1
Prefer not to say       1     1  0           0    0

Answer 3

您可以melt并使用crosstab

df1 = pd.melt(df,id_vars=['ID','Gender'],var_name='Language',value_name='Choice')
df1['Choice'] = np.where(df1['Choice'] == df1['Language'],1,0)
final= pd.crosstab(df1['Gender'],df1['Language'],values=df1['Choice'],aggfunc='sum')
print(final)
Language              Bash  C++  JavaScript  Python  R
Gender                                              
Female                  0    1           1       0  1
Male                    0    0           1       1  0
Prefer not to say       1    0           0       1  0

Answer 4

假设你的nan是NaN(即它不是字符串(，我们可能会利用count，因为它忽略了NaN来获得所需的输出

df_out = df.iloc[:,2:].groupby(df.Gender, sort=False).count()
Out[175]:
Python  Bash  R  JavaScript  C++
Gender
Male                    1     0  0           1    0
Female                  0     0  1           1    1
Prefer not to say       1     1  0           0    0

Answer 5

让我们推到一行

df.drop('ID',1).melt('Gender').
query('variable==value').
groupby(['Gender','variable']).size().unstack(fill_value=0)
Out[120]: 
variable        Bash  C++  JavaScript  Python  R
Gender                                          
Female             0    1           1       0  1
Male               0    0           1       1  0
Prefernottosay     1    0           0       1  0