如何获得分组依据的唯一ID的累计和?

时间:2019-03-07 19:45:04

标签: python pandas data-processing

我对使用python和pandas处理pandas数据框看起来非常陌生

Date     Time           ID   Weight
Jul-1     12:00         A       10
Jul-1     12:00         B       20
Jul-1     12:00         C       100
Jul-1     12:10         C       100
Jul-1     12:10         D       30
Jul-1     12:20         C       100
Jul-1     12:20         D       30
Jul-1     12:30         A       10
Jul-1     12:40         E       40
Jul-1     12:50         F       50
Jul-1     1:00          A       40

我正在尝试按日期,时间和ID进行分组,并应用累加总和,以便如果下一个时隙中存在ID,则权重仅添加一次(唯一)。生成的数据框看起来像这样

Date     Time           Weight   
Jul-1     12:00         130     (10+20+100)
Jul-1     12:10         160     (10+20+100+30)
Jul-1     12:20         160     (10+20+100+30)
Jul-1     12:30         160     (10+20+100+30)
Jul-1     12:40         200     (10+20+100+30+40)
Jul-1     12:50         250     (10+20+100+30+40+50)
Jul-1     01:00         250     (10+20+100+30+40+50)

这是我在下面尝试过的方法,但是仍然多次计算重量:

df=df.groupby(['date','time','ID'])['Wt'].apply(lambda x: x.unique().sum()).reset_index()
df['cumWt']=df['Wt'].cumsum()

任何帮助将不胜感激!

非常感谢!

1 个答案:

答案 0 :(得分:1)

下面的代码使用pandas.duplicate()pandas.merge()pandas.groupby/sumpandas.cumsum()来获得所需的输出:

# creates a series of weights to be considered and rename it to merge
unique_weights = df['weight'][~df.duplicated(['weight'])]
unique_weights.rename('consider_cum', inplace = True)

# merges the series to the original dataframe and replace the ignored values by 0
df = df.merge(unique_weights.to_frame(), how = 'left', left_index=True, right_index=True)
df.consider_cum = df.consider_cum.fillna(0)

# sums grouping by date and time
df = df.groupby(['date', 'time']).sum().reset_index()

# create the cumulative sum column and present the output
df['weight_cumsum'] = df['consider_cum'].cumsum()
df[['date', 'time', 'weight_cumsum']]

产生以下输出:

enter image description here