Question

import pandas as pd
mydate = ["01/01/2018","19/01/2018","24/01/2018" ,
         "27/01/2018","29/01/2018","30/01/2018" , 
         "22/02/2018","23/03/2018"]

mydate = pd.to_datetime(mydate)
events = ["a" , "b" , "c" , "d" , "e" , "f" ,"g" , "h"]

df = pd.DataFrame({"date" :mydate,"events" :events})
df

     date       events
0   2018-01-01  a
1   2018-01-19  b
2   2018-01-24  c
3   2018-01-27  d
4   2018-01-29  e
5   2018-01-30  f
6   2018-02-22  g
7   2018-03-23  h

我想每20天对数据进行切片并将其存储在单独的数据框中。我看了分组，date_range和其他功能，但无法找到我的问题的解决方案。我可以使用典型的for循环来做到这一点，但我希望使用一些pandas功能。

Expected result
df = [df1 , df2 , df3 , df4]
where df1 contain row 0 ,1 
      df2 contains row 2,3,4,5
      df3 contain row 6
      df4 contain row 7

Answer 1

您可以将In [8]: final_list = [e for _, e in df.groupby(pd.Grouper(key='date', freq='20d')) if not e.empty] In [9]: for e in final_list: print(e) date events 0 2018-01-01 a 1 2018-01-19 b date events 2 2018-01-24 c 3 2018-01-27 d 4 2018-01-29 e 5 2018-01-30 f date events 6 2018-02-22 g date events 7 2018-03-23 h与fab.setOnClickListener(new View.OnClickListener() { @Override public void onClick(View view) { Intent startNewPostActivity = new Intent(getApplicationContext(), NewPostActivity.class); startNewPostActivity.putExtra("Fragment_Position", mViewPager.getCurrentItem()); startActivity(startNewPostActivity); } });：

一起使用

{{1}}

Answer 2

这是一个解决方案，虽然它确实使用了一个简单的循环：

import pandas as pd
from datetime import datetime

df = 'your dataframe'

dfs = []

delta = df.date.max() - df.date.min()

for i in range(0, delta.days+1, 20):
     mask = (df['date'] >= df.date.min()+datetime.timedelta(days=i)) & (df['date'] <= df.date.min() + datetime.timedelta(days=i+20))
     dfs.append(df.loc[mask])

Answer 3

我试过了，

minimum=df['date'].min()
df['diff']=(df['date']-minimum)/datetime.timedelta(days=1)

df['s']=df.groupby(pd.cut(df['diff'],np.arange(-0.000001, df['diff'].max()+20, 20))).grouper.group_info[0]
for u,v in df.groupby('s'):
    del v['s']
    print v

输出

        date events  diff
0 2018-01-01      a   0.0
1 2018-01-19      b  18.0
        date events  diff
2 2018-01-24      c  23.0
3 2018-01-27      d  26.0
4 2018-01-29      e  28.0
5 2018-01-30      f  29.0
        date events  diff
6 2018-02-22      g  52.0
        date events  diff
7 2018-03-23      h  81.0

使用时间序列数据将单个Pandas DataFrame拆分为Python中的N DataFrame

3 个答案: