如何利用匹配df['keys']的差异并为它们创建新列

Question

我试图找出给定专业的性别之间的工资差距

以下是我的表格的文本版本：

    gender   field group   logwage
0     male      BUSINESS  7.229572
10  female      BUSINESS  7.072464
1     male    COMM/JOURN  7.108538
11  female    COMM/JOURN  7.015018
2     male  COMPSCI/STAT  7.340410
12  female  COMPSCI/STAT  7.169401
3     male     EDUCATION  6.888829
13  female     EDUCATION  6.770255
4     male   ENGINEERING  7.397082
14  female   ENGINEERING  7.323996
5     male    HUMANITIES  7.053048
15  female    HUMANITIES  6.920830
6     male      MEDICINE  7.319011
16  female      MEDICINE  7.193518
17  female        NATSCI  6.993337
7     male        NATSCI  7.089232
18  female         OTHER  6.881126
8     male         OTHER  7.091698
9     male  SOCSCI/PSYCH  7.197572
19  female  SOCSCI/PSYCH  6.968322

diff不适合我，因为它会在每个连续的专业之间产生差异

下面是目前的代码：

for row in sorted_mfield:
   if sorted_mfield['field group']==sorted_mfield['field group'].shift(1):
    diff= lambda x: x[0]-x[1]

我的下一个策略是回到未排序的数据框架，其中男性和女性是他们自己的专栏，并从中有所不同，但由于我已经花了一个小时尝试这样做，而且我对熊猫非常陌生，我想我会询问并了解这是如何工作的。谢谢

Answer 1

我会考虑用^ {CD1>}重塑您的数据文件，使计算变得更容易。

代码：

df.pivot(index='field group', columns='gender', values='logwage').rename_axis([None], axis=1)

#                female      male
#field group                     
#BUSINESS      7.072464  7.229572
#COMM/JOURN    7.015018  7.108538
#COMPSCI/STAT  7.169401  7.340410
#EDUCATION     6.770255  6.888829
#ENGINEERING   7.323996  7.397082
#HUMANITIES    6.920830  7.053048
#MEDICINE      7.193518  7.319011
#NATSCI        6.993337  7.089232
#OTHER         6.881126  7.091698
#SOCSCI/PSYCH  6.968322  7.197572

df.male - df.female

#field group
#BUSINESS        0.157108
#COMM/JOURN      0.093520
#COMPSCI/STAT    0.171009
#EDUCATION       0.118574
#ENGINEERING     0.073086
#HUMANITIES      0.132218
#MEDICINE        0.125493
#NATSCI          0.095895
#OTHER           0.210572
#SOCSCI/PSYCH    0.229250
#dtype: float64

Answer 2

在数据的排序版本中使用Pandas.DataFrame.shift()的解决方案：

df.sort_values(by=['field group', 'gender'], inplace=True)
df['gap'] = df.logwage - df.logwage.shift(1)
df[df.gender =='male'][['field group', 'gap']]

使用示例数据生成以下输出：

    field group     gap
0   BUSINESS        0.157108
2   COMM/JOURN      0.093520
4   COMPSCI/STAT    0.171009
6   EDUCATION       0.118574
8   ENGINEERING     0.073086
10  HUMANITIES      0.132218
12  MEDICINE        0.125493
15  NATSCI          0.095895
17  OTHER           0.210572
18  SOCSCI/PSYCH    0.229250

注意：它认为每个字段组始终有一对值。如果要验证它或消除没有此对的字段组，请使用下面的代码进行筛选：

df_grouped = df.groupby('field group') 
df_filtered = df_grouped.filter(lambda x: len(x) == 2)