Question

我有dataframe，我需要按“ sev”列按条件1＆2，3、4＆5进行分组并找到其计数。

有什么方法吗？

我对此感到厌倦，但是在sev列中为每个单独的值给出了

import os
import requests
from time import time
import uuid
from multiprocessing.pool import ThreadPool
main_file_name = 'test1.csv'

my_set = set()
with open(main_file_name, 'r') as f:  #read image urls 
    for row in f:
        my_set.add(row.split(',')[2].strip())

def get_url(entry):
    path = str(uuid.uuid4()) + ".jpg"
    if not os.path.exists(path):
        r = requests.get(entry, stream=True)
        if r.status_code == 200:
            with open(path, 'wb') as f:
                for chunk in r:
                    f.write(chunk)

start = time()
results = ThreadPool(8).imap_unordered(get_url, my_set)
print(f"Elapsed Time: {time() - start}")

pandas dataFrame-

df.groupby(['sev']).ids.agg('count').to_frame('count').reset_index()

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}})

预期产量

| 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   |

Answer 1

主要问题是您需要将严重性级别汇总为较少的类别，这可以通过pd.cut来完成，因为sev是数字，并且您希望以连续的间隔进行。如果不是数字或间隔不连续（例如1＆4、2、3＆5），则需要df.replace和映射字典。

然后，可以使用df.pivot_table或通过groupby/unstack“手动”完成重塑。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

熊猫按条件按列值分组

1 个答案: