如何为“检票口数量”创建一列答案

【问题标题】：How to create a column for 'number of wickets'如何为“检票口数量”创建一列
【发布时间】：2021-12-09 19:39:49
【问题描述】：

它应该是什么样子的示例：

这是上表的输出：

{'batfast_id': {0: 'bfs1',
  1: 'bfs1',
  2: 'bfs1',
  3: 'bfs1',
  4: 'bfs1',
  5: 'bfs1',
  6: 'bfs1',
  7: 'bfs2',
  8: 'bfs2',
  9: 'bfs2',
  10: 'bfs2',
  11: 'bfs2',
  12: 'bfs2'},
 'score': {0: 1,
  1: 2,
  2: 0,
  3: 0,
  4: 1,
  5: 0,
  6: 3,
  7: 3,
  8: 0,
  9: 0,
  10: 2,
  11: 2,
  12: 0},
 'day_month_year': {0: '01-01-2020',
  1: '01-01-2020',
  2: '01-01-2020',
  3: '01-01-2020',
  4: '02-01-2020',
  5: '02-01-2020',
  6: '02-01-2020',
  7: '02-01-2020',
  8: '02-01-2020',
  9: '02-01-2020',
  10: '03-01-2020',
  11: '03-01-2020',
  12: '03-01-2020'},
 'runs': {0: 1,
  1: 3,
  2: 3,
  3: 3,
  4: 1,
  5: 1,
  6: 4,
  7: 3,
  8: 3,
  9: 3,
  10: 2,
  11: 4,
  12: 4},
 'deliveries_faced': {0: 1,
  1: 2,
  2: 3,
  3: 4,
  4: 1,
  5: 2,
  6: 3,
  7: 1,
  8: 2,
  9: 3,
  10: 1,
  11: 2,
  12: 3},
 'wicket': {0: 'Not Out',
  1: 'Not Out',
  2: 'Bowled',
  3: 'Caught',
  4: 'Not Out',
  5: 'Caught',
  6: 'Not Out',
  7: 'Not Out',
  8: 'Bowled',
  9: 'Bowled',
  10: 'Not Out',
  11: 'Not Out',
  12: 'Caught'},
 'no_of_wickets': {0: 0,
  1: 0,
  2: 1,
  3: 2,
  4: 0,
  5: 1,
  6: 1,
  7: 0,
  8: 1,
  9: 2,
  10: 0,
  11: 0,
  12: 1}}

我正在尝试创建一个名为 no_of_wickets 的列，用于计算用户在打板球时失去了多少个三柱门。但是，当日期更改或batfast_id（用户）更改时，它必须重置为 0。

score 是玩家在该次投球中获得的跑动次数，runs 是score 的累积次数。

runs 是使用以下代码创建的：df['runs']=df.groupby(['batfast_id','day_month_year'])['score'].cumsum()

deliveries_faced 使用此代码：df['deliveries_faced']=df.groupby(['batfast_id','day_month_year']).cumcount()+1

【问题讨论】：

标签： python pandas pandas-groupby cumsum

【解决方案1】：

创建一个布尔掩码，如果wicket 不在，则设置为True，然后按batfast_id 和day_month_year 分组，最后计算累积总和。

df['no_of_wickets'] = df.assign(is_out=df['wicket'].ne('Not Out')) \
                        .groupby(['batfast_id', 'day_month_year'])['is_out'] \
                        .cumsum()
print(df)

# Output:
   batfast_id  score day_month_year  runs  deliveries_faced   wicket  no_of_wickets
0        bfs1      1     01-01-2020     1                 1  Not Out              0
1        bfs1      2     01-01-2020     3                 2  Not Out              0
2        bfs1      0     01-01-2020     3                 3   Bowled              1
3        bfs1      0     01-01-2020     3                 4   Caught              2
4        bfs1      1     02-01-2020     1                 1  Not Out              0
5        bfs1      0     02-01-2020     1                 2   Caught              1
6        bfs1      3     02-01-2020     4                 3  Not Out              1
7        bfs2      3     02-01-2020     3                 1  Not Out              0
8        bfs2      0     02-01-2020     3                 2   Bowled              1
9        bfs2      0     02-01-2020     3                 3   Bowled              2
10       bfs2      2     03-01-2020     2                 1  Not Out              0
11       bfs2      2     03-01-2020     4                 2  Not Out              0
12       bfs2      0     03-01-2020     4                 3   Caught              1

【讨论】：

是的，已经成功了，非常感谢您的帮助！