带有条件的pandas by列值
带有条件的pandas by列值
提问于 2020-09-28 02:11:01
我有数据帧，我需要按“sev”列分组，条件是1&2，3，4&5，并找到它的计数。
有什么方法吗？
我已经厌倦了这一点，但它给了sev列中每个单独的值
df.groupby(['sev']).ids.agg('count').to_frame('count').reset_index()pandas dataFrame -
df = pd.DataFrame({'ids': {0: 'D1791272223', 1: 'V25369085223', 2: 'V25117230523', 3: 'V25104327323', 4: 'V24862169823', 5: 'P3944221523', 6: 'V24776335823', 7: 'V24722584123', 8: 'V24716191923', 9: 'V24575876123', 10: 'V24791923'}, 'status': {0: 'Resolved', 1: 'Resolved', 2: 'Resolved', 3: 'Resolved', 4: 'Open', 5: 'Open', 6: 'Closed', 7: 'Resolved', 8: 'Resolved', 9: 'Open', 10: 'Resolved'}, 'action': {0: 'Comment', 1: 'Implementation', 2: 'Comment', 3: 'Implementation', 4: 'Comment', 5: 'Implementation', 6: 'Comment', 7: 'Comment', 8: 'Implementation', 9: 'Comment', 10: 'Implementation'}, 'sev': {0: 3, 1: 2, 2: 1, 3: 3, 4: 4, 5: 4, 6: 3, 7: 2, 8: 2, 9: 1, 10: 5}})| ids | status | action | sev |
|--------------|----------|----------------|-----|
| D1791272223 | Resolved | Comment | 3 |
| V25369085223 | Resolved | Implementation | 2 |
| V25117230523 | Resolved | Comment | 1 |
| V25104327323 | Resolved | Implementation | 3 |
| V24862169823 | Open | Comment | 4 |
| P3944221523 | Open | Implementation | 4 |
| V24776335823 | Closed | Comment | 3 |
| V24722584123 | Resolved | Comment | 2 |
| V24716191923 | Resolved | Implementation | 2 |
| V24575876123 | Open | Comment | 1 |
| V24791923 | Resolved | Implementation | 5 |预期输出
| sev | count | Open count | Closed and Resolved count |
|--------|-------|------------|-----------------------------|
| 1&2 | 5 | 1 | 4 |
| 3 | 3 | 0 | 3 |
| 4&5 | 3 | 2 | 1 |回答 1
Stack Overflow用户
回答已采纳
发布于 2020-09-28 02:44:59
主要问题是，您需要将严重性级别聚合到较少的类别中，这可以使用pd.cut完成，因为sev是数字，并且您希望按连续的时间间隔进行。如果它不是数字或间隔不连续(例如1&4、2、3和5)，您将需要具有映射字典的df.replace。
然后，可以使用df.pivot_table或使用groupby/unstack“手动”完成整形。我更喜欢groupby，因为它在其他情况下更灵活。
df['sev_group'] = pd.cut(df['sev'], bins=[0, 2, 3, 5],
labels=['1&2', '3', '4&5'])
summary = df.groupby(['sev_group', 'status']).size().unstack()
# or
# summary = df.pivot_table(values='ids', index='sev_group',
# columns='status', aggfunc='count', fill_value=0)
summary['count'] = summary.sum(axis=1)
summary['Closed/Resolved'] = summary['Closed'] + summary['Resolved']
summary = summary[['count', 'Open', 'Closed/Resolved']]输出
status count Open Closed/Resolved
sev_group
1&2 5 1 4
3 3 0 3
4&5 3 2 1页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/64091744
复制相关文章
相似问题
腾讯云开发者
