Python 将pd datetime对象更改为整数

我有一个熊猫数据框，里面有两个日期。我想计算他们之间的天数差。但由此产生的差异看起来像字符串ex（“7天”）。有没有办法将此更改为整数日期差

y['datePulled'] = pd.to_datetime(y['datePulled'])
y['Dates'] = pd.to_datetime(y['Dates'])
y['Datediff'] = y['datePulled'] - y['Dates']
y['Datediff']
0    7 days
1    6 days
2    5 days
3    4 days
4    3 days
5    2 days
6    1 days

您可以使用：

(y['Datediff'] / np.timedelta64(1, 'D')).astype(int)

或：

样本：

import pandas as pd
import numpy as np

y = pd.DataFrame({ 'datePulled': ['2016-01-05','2016-01-04'], 
                    'Dates': ['2016-01-01','2016-01-02']})

y['datePulled'] = pd.to_datetime(y['datePulled'])
y['Dates'] = pd.to_datetime(y['Dates'])
y['Datediff'] = y['datePulled'] - y['Dates']
print (y)

#output is float, cast to int
y['Datediff1'] = (y['Datediff'] / np.timedelta64(1, 'D')).astype(int)

y['Datediff2'] = y['Datediff'].dt.days
print (y)
       Dates datePulled  Datediff  Datediff1  Datediff2
0 2016-01-01 2016-01-05    4 days          4          4
1 2016-01-02 2016-01-04    2 days          2          2

在较大的数据帧中，第一种方法速度更快：

y = pd.concat([y]*1000).reset_index(drop=True)

In [236]: %timeit (y['Datediff'] / np.timedelta64(1, 'D')).astype(int)
1000 loops, best of 3: 789 µs per loop

In [237]: %timeit y['Datediff'].dt.days
100 loops, best of 3: 15.3 ms per loop

---

获取以下错误：回溯（最近一次调用last）：文件“”，第1行，类型错误：元数据中的日期时间单位“d”无效

y = pd.concat([y]*1000).reset_index(drop=True)

In [236]: %timeit (y['Datediff'] / np.timedelta64(1, 'D')).astype(int)
1000 loops, best of 3: 789 µs per loop

In [237]: %timeit y['Datediff'].dt.days
100 loops, best of 3: 15.3 ms per loop