每1000倍重置Pandas Cumsum

Question

目前，我有一个如下所示的数据格式，每次它跨越1000ex (2000,3000...etc)的倍数时，我都需要重置累计和。

                    Production    ID  cumsum  
     2017-10-19        1054  1323217    1054     
     2017-10-20           0  1323217    1054     
     2017-10-21           0  1323217    1054     
     2017-10-22           0  1323217    1054     
     2017-10-23           0  1323217    1054  

例如，在上面，我需要一个如下所示的df：

                 Production    ID      cumsum  adjCumsum numberGenerated
      2017-10-19        1054  1323217    1054     1000      1
      2017-10-20           0  1323217    1054     54        0
      2017-10-21           0  1323217    1054     54        0
      2017-10-22        3054  1323217    4108     4000      4
      2017-10-23           0  1323217    4018     108       0 
      2017-10-23         500  1323218    500      500       0

下面，每1000正确重置一次值，但我似乎不太明白如何通过ID对其进行分组，并将其舍入到1000 s。

maxvalue = 1000

lastvalue = 0
newcum = []
    for row in df.iterrows():
        thisvalue =  row[1]['cumsum'] + lastvalue
           if thisvalue > maxvalue:
              thisvalue = 0
           newcum.append( thisvalue )
           lastvalue = thisvalue
  df['newcum'] = newcum

由于下面的答案，我现在能够计算生成的累积数，但我需要计算生成的增量#。

     df['cumsum'] = df.groupby('ID')['Production'].cumsum()
     thresh = 1000
     multiple = (df['cumsum'] // thresh )
     mask = multiple.diff().ne(0)
     df['numberGenerated'] = np.where(mask, multiple, 0)
     df['adjCumsum'] = (df['numberGenerated'].mul(thresh)) + df['cumsum'] % 
     thresh

    df['cumsum2'] = df.groupby('ID')['numberGenerated'].cumsum()

My initial thinking was to try something similar to:

      df['numGen1'] = df['cumsum2'].diff()

最终编辑测试和工作。谢谢你的帮助，

I was overthinking it, below is how I was able to do it:

df['cumsum'] = df.groupby('ID')['Production'].cumsum()
thresh = 1000

multiple = (df['cumsum'] // thresh )

mask = multiple.diff().ne(0)
df['numberGenerated'] = np.where(mask, multiple, 0)
df['adjCumsum'] = (df['numberGenerated'].mul(thresh)) + df['cumsum'] % thresh

df['cumsum2'] = df.groupby('ID')['numberGenerated'].cumsum()

numgen = []
adjcumsum = []

for i in range(len(df['cumsum'])):
    if df['cumsum'][i] > thresh and (df['ID'][i] == df['ID'][i-1]):
        numgenv = (df['cumsum'][i] // thresh) - (df['cumsum'][i-1] // thresh)
        numgen.append(numgenv)
    elif df['cumsum'][i] > thresh:
        numgenv = (df['cumsum'][i] // thresh)
        numgen.append(numgenv)
    else:
        numgenv = 0
        numgen.append(numgenv)

df['numgen2.0'] = numgen

Answer 1

IIUC，这只是一个整数除法问题，有一些技巧：

thresh = 1000
df['cumsum'] = df['Production'].cumsum()

# how many times cumsum passes thresh
multiple = (df['cumsum'] // thresh )

# detect where thresh is pass
mask = multiple.diff().ne(0)

# update the number generated:
df['numberGenerated'] = np.where(mask, multiple, 0)

# then the adjusted cumsum 
df['adjCumsum'] = (df['numberGenerated'].mul(thresh)) + df['cumsum'] % thresh

输出：

            Production       ID  cumsum  adjCumsum  numberGenerated
2017-10-19        1054  1323217    1054       1054                1
2017-10-20           0  1323217    1054         54                0
2017-10-21           0  1323217    1054         54                0
2017-10-22        3054  1323217    4108       4108                4
2017-10-23           0  1323217    4108        108                0
2017-10-23         500  1323218    4608        608                0