pandas 将值与前一行与过滤条件进行比较

问题描述

我有一个包含员工工资信息的 DataFrame.大约有 900000+ 行.

I have a DataFrame with information about employee salary. It's about 900000+ rows.

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

我需要添加一列，其中包含有关增加减少的工资水平的信息.我正在使用 shift() 函数来比较行中的值.

I need to add a column with information about increaseddecreased salary level. I'm using the shift() function to compare value in rows.

主要问题在于对整个数据集的所有唯一员工进行过滤和迭代.

Main problem is in filtration and iteration over all unique employees over the whole dataset.

在我的脚本中大约需要 3 个半小时.

如何做到更快?

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.]})

样本输出:

+----+-------------+---------------+----------+-------+
|    |   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 |
+----+-------------+---------------+----------+-------+

推荐答案

使用 groupby 和 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

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