python - 如果 DST 更改的数据帧频率低于 1 小时，则会出现 pytz 错误 [多索引 pandas]

Question

我在更改频率低于 1 小时的数据帧的时区时遇到问题。就我而言，我从 CSV 源获取每刻钟一次的数据帧，并且必须删除 3 月份的 DST 时间并添加 10 月份的 DST 时间。如果频率是每小时，下面的函数可以很好地工作，但是如果频率低于以下的频率，下面的函数就不起作用。

有人能解决这个问题吗？

import pandas as pd
import numpy as np
from pytz import timezone

def DST_Paris(NH, NH_str):
    ## Suppose that I do not create the dataframe here but I import one from a CSV file
    df = pd.DataFrame(np.random.randn(NH * 365), index = pd.date_range(start="01/01/2014", freq=NH_str, periods=NH * 365))

    ## I need to delete the hour in March and duplicate the hour in October
    ## If freq is inf at 1 hour, I need to duplicate all the data inside the considerated hour 

    tz = timezone('Europe/Paris')
    change_date = tz._utc_transition_times
    GMT1toGMT2_dates = [datei.date() for datei in list(change_date) if datei.month == 3]
    GMT2toGMT1_dates = [datei.date() for datei in list(change_date) if datei.month == 10] 
    ind_March   = np.logical_and(np.in1d(df.index.date, GMT1toGMT2_dates),(df.index.hour == 2))
    ind_October = np.logical_and(np.in1d(df.index.date, GMT2toGMT1_dates),(df.index.hour == 2))
    df['ind_March'] = (1-ind_March)
    df['ind_October'] = ind_October * 1
    df = df[df.ind_March == 1]
    df = df.append(df[df.ind_October == 1])
    del df['ind_March']
    del df['ind_October']
    df = df.sort()

    ## Error if granularity below of 1 hours
    df = df.tz_localize('Europe/Paris', ambiguous = 'infer')
    return df

try:
    DST_Paris(24, "1h")
    print "dataframe freq = 1h    ==> no pb"
except:
    print "dataframe freq = 1h    ==> error"

try:
    DST_Paris(96, "15min")
    print "dataframe freq = 15min ==> no pb"
except:
    print "dataframe freq = 15min ==> error"

输出是:

dataframe freq = 1h    ==> no pb
dataframe freq = 15min ==> error

Answer 1

解决方法是使用

is_dst = False  # or True
df = df.tz_localize('Europe/Paris', ambiguous=[is_dst]*len(df))

明确指定是否应将不明确的本地时间解释为夏令时区。

<小时/>

顺便说一下，

df['ind_March'] = (1-ind_March)
df['ind_October'] = ind_October * 1
df = df[df.ind_March == 1]
df = df.append(df[df.ind_October == 1])
del df['ind_March']
del df['ind_October']
df = df.sort()

可以简化为

df = df.loc[(~ind_March) & (ind_October)] 
df = df.sort()