Question

我正在处理这个问题几个小时。这一定是一个小小的修复，但是我不知何故是盲目的。

This thread不能解决我的问题。

这是我的数据

Date        Server
2019-02-13  A
2019-02-13  B
2019-02-13  B
2019-02-17  A
2019-02-17  B
2019-02-17  C
2019-02-19  C
2019-02-19  D

我需要获取相应日期范围内的服务器列表。我尝试了这段代码：

df['Date'] = pd.to_datetime(df['Date'], format='%Y%m%d').apply(lambda x: x.strftime(format='%Y-%m-%d'))

df = df.set_index(df['Date'])

### This formatting changes the cell content from a format like 20190217 to the 
one represented above. Maybe there is already an error right here.### 

start_date = pd.to_datetime('20190212', format='%Y%m%d').strftime(format='%Y-%m-%d')
end_date   = pd.to_datetime('20190217', format='%Y%m%d').strftime(format='%Y-%m-%d')

但是，如果我明确地写出日期，则打印语句将提供正确的结果。但是，在我的程序中，我需要按开始日期和结束日期输入日期。

print(df[df.Date.between('2019-02-12','2019-02-17')].Server.unique())
print(df.loc['2019-02-12':'2019-02-17'].Server.unique())
print(df.loc[start_date : end_date].Server.unique())

输出：

['A' 'B' 'C']     - correct
['A' 'B' 'C']     - correct
['A' 'B' 'C' 'D'] - incorrect

我需要对代码进行哪些更改？

谢谢您的帮助！

Answer 1

您无需制作strftime并将格式更改为format='%Y-%m-%d'

import pandas as pd

df = pd.DataFrame({'Date': ['2019-02-13', '2019-02-13', '2019-02-13', '2019-02-17', '2019-02-17', '2019-02-17', '2019-02-19', '2019-02-19'],
                   'Server':['A','B','B','A','B','C','C','D']})


df['Date'] = pd.to_datetime(df['Date'], format='%Y-%m-%d')
df = df.set_index(df['Date'])
start_date = pd.to_datetime('20190212', format='%Y%m%d').strftime(format='%Y-%m-%d')
end_date   = pd.to_datetime('20190217', format='%Y%m%d').strftime(format='%Y-%m-%d')
print(df[df.Date.between('2019-02-12','2019-02-17')].Server.unique())
print(df.loc['2019-02-12':'2019-02-17'].Server.unique())
print(df.loc[start_date : end_date].Server.unique())

输出为

['A' 'B' 'C']
['A' 'B' 'C']
['A' 'B' 'C']

Answer 2

这应该可以解决问题。

import pandas as pd
start_date = '2019-02-12'
end_date = '2019-02-17'
df['Date'] = pd.to_datetime(df['Date'])
print(df.loc[(df['Date'] > start_date) & (df['Date'] <= end_date)].Server.unique())

按日期范围选择DataFrame行

2 个答案: