熊猫在一系列中迭代可变长度切片的最大值答案

【问题标题】：Pandas iterate max value of a variable length slice in a series熊猫在一系列中迭代可变长度切片的最大值
【发布时间】：2017-10-20 00:19:56
【问题描述】：

假设我有一个 Pandas DataFrame，如下所示：

import pandas as pd
idx = ['2003-01-02', '2003-01-03', '2003-01-06', '2003-01-07',
       '2003-01-08', '2003-01-09', '2003-01-10', '2003-01-13',
       '2003-01-14', '2003-01-15', '2003-01-16', '2003-01-17',
       '2003-01-21', '2003-01-22', '2003-01-23', '2003-01-24',
       '2003-01-27']

a = pd.DataFrame([1,2,0,0,1,2,3,0,0,0,1,2,3,4,5,0,1],
                  columns = ['original'], index = pd.to_datetime(idx))

我试图在两个零之间获取该 DataFrame 的每个切片的最大值。在那个例子中，我会得到：

a['result'] = [0,2,0,0,0,0,3,0,0,0,0,0,0,0,5,0,1]

即：

            original  result
2003-01-02         1       0
2003-01-03         2       2
2003-01-06         0       0
2003-01-07         0       0
2003-01-08         1       0
2003-01-09         2       0
2003-01-10         3       3
2003-01-13         0       0
2003-01-14         0       0
2003-01-15         0       0
2003-01-16         1       0
2003-01-17         2       0
2003-01-21         3       0
2003-01-22         4       0
2003-01-23         5       5
2003-01-24         0       0
2003-01-27         1       1

【问题讨论】：

标签： pandas python-3.5

【解决方案1】：

找零
cumsum组群
mask 零成自己的组-1
在每个组中找到最大位置idxmax
去掉-1 组的那个，反正那个是零
获取 a.original 以获取找到的最大位置，重新索引并用零填充

m = a.original.eq(0)
g = a.original.groupby(m.cumsum().mask(m, -1))
i = g.idxmax().drop(-1)
a.assign(result=a.loc[i, 'original'].reindex(a.index, fill_value=0))

            original  result
2003-01-02         1       0
2003-01-03         2       2
2003-01-06         0       0
2003-01-07         0       0
2003-01-08         1       0
2003-01-09         2       0
2003-01-10         3       3
2003-01-13         0       0
2003-01-14         0       0
2003-01-15         0       0
2003-01-16         1       0
2003-01-17         2       0
2003-01-21         3       0
2003-01-22         4       0
2003-01-23         5       5
2003-01-24         0       0
2003-01-27         1       1

【讨论】：