Question

我有一个DataFrame，其中包含有关员工薪水的信息。大约有900000多行。

示例：

+----+-------------+---------------+----------+
|    |   table_num | name          |   salary |
|----+-------------+---------------+----------|
|  0 |      001234 | John Johnson  |     1200 |
|  1 |      001234 | John Johnson  |     1000 |
|  2 |      001235 | John Johnson  |     1000 |
|  3 |      001235 | John Johnson  |     1200 |
|  4 |      001235 | John Johnson  |     1000 |
|  5 |      001235 | Steve Stevens |     1000 |
|  6 |      001236 | Steve Stevens |     1200 |
|  7 |      001236 | Steve Stevens |     1200 |
|  8 |      001236 | Steve Stevens |     1200 |
+----+-------------+---------------+----------+

dtypes：

table_num: string
name: string
salary: float

我需要添加一列有关薪资水平提高/降低的信息。我正在使用shift()函数比较行中的值。

主要问题在于整个数据集中所有唯一雇员的过滤和迭代。

我的脚本大约需要3个半小时。

如何更快地做到这一点？

我的脚本：

# giving us only unique combination of 'table_num' and 'name'
    # since there can be same 'table_num' for different 'name'
    # and same names with different 'table_num' appears sometimes

names_df = df[['table_num', 'name']].drop_duplicates()

# then extracting particular name and table_num from Series
for i in range(len(names_df)):    ### Bottleneck of whole script ###    
    t = names_df.iloc[i,[0,1]][0]
    n = names_df.iloc[i,[0,1]][1]

    # using shift() and lambda to check if there difference between two rows 
    diff_sal = (df[(df['table_num']==t)
               & ((df['name']==n))]['salary'] - df[(df['table_num']==t)
                                                 & ((df['name']==n))]['salary'].shift(1)).apply(lambda x: 1 if x>0 else (-1 if x<0 else 0))
    df.loc[diff_sal.index, 'inc'] = diff_sal.values

样本输入数据：

df = pd.DataFrame({'table_num': ['001234','001234','001235','001235','001235','001235','001236','001236','001236'], 
                     'name': ['John Johnson','John Johnson','John Johnson','John Johnson','John Johnson', 'Steve Stevens', 'Steve Stevens', 'Steve Stevens', 'Steve Stevens'], 
                     'salary':[1200.,1000.,1000.,1200.,1000.,1000.,1200.,1200.,1200.]})

示例输出：

+----+-------------+---------------+----------+-------+
|    |   table_num | name          |   salary |   inc |
|----+-------------+---------------+----------+-------|
|  0 |      001234 | John Johnson  |     1200 |     0 |
|  1 |      001234 | John Johnson  |     1000 |    -1 |
|  2 |      001235 | John Johnson  |     1000 |     0 |
|  3 |      001235 | John Johnson  |     1200 |     1 |
|  4 |      001235 | John Johnson  |     1000 |    -1 |
|  5 |      001235 | Steve Stevens |     1000 |     0 |
|  6 |      001236 | Steve Stevens |     1200 |     0 |
|  7 |      001236 | Steve Stevens |     1200 |     0 |
|  8 |      001236 | Steve Stevens |     1200 |     0 |
+----+-------------+---------------+----------+-------+

Answer 1

与groupby一起使用diff：

df['inc'] = df.groupby(['table_num', 'name'])['salary'].diff().fillna(0.0)
df.loc[df['inc'] > 0.0, 'inc'] = 1.0
df.loc[df['inc'] < 0.0, 'inc'] = -1.0

Answer 2

将DataFrameGroupBy.diff与numpy.sign一起使用，并最后投射到integer s：

df['new'] = np.sign(df.groupby(['table_num', 'name'])['salary'].diff().fillna(0)).astype(int)
print (df)
   table_num           name  salary  new
0       1234   John Johnson    1200    0
1       1234   John Johnson    1000   -1
2       1235   John Johnson    1000    0
3       1235   John Johnson    1200    1
4       1235   John Johnson    1000   -1
5       1235  Steve Stevens    1000    0
6       1236  Steve Stevens    1200    0
7       1236  Steve Stevens    1200    0
8       1236  Steve Stevens    1200    0

Answer 3

shift()是解决之道，但您应尽可能避免使用循环。在这里，我们可以利用groupby()和transform()的力量。检查熊猫docs。

您可以通过以下方式做到：

df.assign(inc=lambda x: x.groupby(['name'])
                      .salary
                      .transform(lambda y: y - y.shift(1))
                      .apply(lambda x: 1 if x>0 else (-1 if x<0 else 0))
      )

产量：

    table_num   name       salary   inc
0   001234  John Johnson    1200.0  0
1   001234  John Johnson    1000.0  -1
2   001235  John Johnson    1000.0  0
3   001235  John Johnson    1200.0  1
4   001235  John Johnson    1000.0  -1
5   001235  Steve Stevens   1000.0  0
6   001236  Steve Stevens   1200.0  1
7   001236  Steve Stevens   1200.0  0
8   001236  Steve Stevens   1200.0  0

Answer 4

我认为您可以搜索以下术语：“熊猫矢量化”以加快数据框的操作速度，对于您的问题，您可以尝试以下操作：

import pandas as pd

df = pd.DataFrame({'table_num': ['001234','001234','001235','001235','001235','001235','001236','001236','001236'],
                     'name': ['John Johnson','John Johnson','John Johnson','John Johnson','John Johnson', 'Steve Stevens', 'Steve Stevens', 'Steve Stevens', 'Steve Stevens'],
                     'salary':[1200.,1000.,1000.,1200.,1000.,1000.,1200.,1200.,1200.]})

df['temp'] = df['name'] + df['table_num']
df.sort_values('temp', inplace=True)
df['diff'] = df.groupby('temp')['salary'].diff()
df['diff'] = (df['diff'] / abs(df['diff'])).fillna(0)

熊猫将条件与上一行进行比较

4 个答案: