python - Pandas 将值与具有过滤条件的前一行进行比较

Question

我有一个包含员工薪水信息的 DataFrame。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 |
+----+-------------+---------------+----------+

数据类型:

table_num: string
name: string
salary: float

我需要添加一个列，其中包含有关增加\减少工资水平的信息。我正在使用 shift() 函数来比较行中的值。

主要问题在于过滤和迭代整个数据集中的所有唯一员工。

我的脚本大约需要 3 个半小时。

如何更快地完成？

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

示例输入数据:

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

使用groupby连同 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