如何摆脱 MonthEnds 类型

Question

我正在尝试获取 Pandas DataFrame 中开始日期和结束日期之间的月差值。结果并不完全令人满意......

首先，结果是某种格式为 <[value] * MonthEnds> 的 Datetime 类型。我不能用这个来计算。第一个问题是如何将其转换为整数。我尝试了 .n 属性，但随后出现以下错误：

AttributeError: 'Series' object has no attribute 'n'  

二、结局是“失踪”一个月。这可以通过使用其他解决方案/方法来避免吗？或者我应该只在答案上加 1 个月？

为了支持我的问题，我创建了一些简化的代码：

dates = [{'Start':'1-1-2020', 'End':'31-10-2020'}, {'Start':'1-2-2020', 'End':'30-11-2020'}]

df = pd.DataFrame(dates)

df['Start'] = pd.to_datetime(df['Start'], dayfirst=True)

df['End'] = pd.to_datetime(df['End'], dayfirst=True)

df['Duration'] = (df['End'].dt.to_period('M') - df['Start'].dt.to_period('M'))

df

这导致：

    Start       End         Duration

0   2020-01-01  2020-10-31  <9 * MonthEnds>

1   2020-02-01  2020-11-30  <9 * MonthEnds>

首选结果是：

    Start       End         Duration

0   2020-01-01  2020-10-31  10

1   2020-02-01  2020-11-30  10

Answer 1

从结束日期中减去开始日期并将时间增量转换为月份。import pandas as pddates = [{'Start':'1-1-2020', 'End':'31-10-2020'}, {'Start':'1-2-2020', 'End':'30-11-2020'}]df = pd.DataFrame(dates)df['Start'] = pd.to_datetime(df['Start'], dayfirst=True)df['End'] = pd.to_datetime(df['End'], dayfirst=True)df['Duration'] = (df['End']-df['Start']).astype('<m8[M]').astype(int)+1print(df)输出：&nbsp; &nbsp; &nbsp; &nbsp;Start&nbsp; &nbsp; &nbsp; &nbsp; End&nbsp; Duration0 2020-01-01 2020-10-31&nbsp; &nbsp; &nbsp; &nbsp; 101 2020-02-01 2020-11-30&nbsp; &nbsp; &nbsp; &nbsp; 10

Answer 2

尝试这个dates = [{'Start':'1-1-2020', 'End':'31-10-2020'}, {'Start':'1-2-2020', 'End':'30-11-2020'}]df = pd.DataFrame(dates)df['Start'] = pd.to_datetime(df['Start'], dayfirst=True)df['End'] = pd.to_datetime(df['End'], dayfirst=True)df['Duration'] = (df['End'] - df['Start']).apply(lambda x:x.days//30)print(df)