python:将列转换为日期格式并提取顺序

听起来你有两个问题。

cols = [col for col in df.columns if col != 'ID']
df.loc[:, cols] = df.loc[:, cols].applymap(lambda x: datetime.strptime('1582/10/15', "%Y/%m/%d") + timedelta(days=x) if np.isfinite(x) else x)

2)要获取排序的列名:

df['ORDER'] = df.loc[:, cols].apply(lambda dr: ','.join(df.loc[:, cols].columns[dr.dropna().argsort()]), axis=1)

首先,我将使输入列以逗号分隔,以便更容易处理表单:

ID,ACT1,ACT2,ACT3,ACT4
1,154438.0,154104.0,155321.0,155321.0
2,154042.0,154073.0,154104.0,154104.0

#!/usr/bin/env python3

import csv
from operator import itemgetter

idAndTuple = {}
with open('time.txt') as csvfile:
  reader = csv.DictReader(csvfile)
  for row in reader:
    myID = row.pop('ID',None)
    reorderedList = sorted(row.items(), key = itemgetter(1))
    idAndTuple[myID] = reorderedList
    print( myID, reorderedList )

1 [('ACT2', '154104.0'), ('ACT1', '154438.0'), ('ACT3', '155321.0'), ('ACT4', '155321.0')]
2 [('ACT1', '154042.0'), ('ACT2', '154073.0'), ('ACT3', '154104.0'), ('ACT4', '154104.0')]