Python_时间类型转换

[TOC]

1. 时间转换

对于时间序列的转换，好像都是通过datetime作为中转来变换timestamp和字符串的，也就是：

timestamp ←→ datetime ←→ str

timestamp  → datetime  pd.datetime.fromtimestamp(1487312878)
str        → datetime  pd.to_datetime('2017-01-02')

import datetime
datetime 变量a  →  timestamp   a.timestamp()
datetime 变量a  →  str         a.strftime('%Y-%m-%d')

pandas应用：

timestamp  → datetime  Seriesobj.map(pd.datetime.fromtimestamp)
str        → datetime  Seriesobj.map(pd.to_datetime)

datetime 变量x  →  timestamp   Seriesobj.map(lambda x.timestamp()).astype(np.int64)
datetime 变量x  →  str         Seriesobj.map(lambda x.strftime('%Y-%m-%d %H-%m')) 

2. 计算2个日期之间的时间差

best-way-to-find-the-months-between-two-dates

Python: Difference of 2 datetimes in months [duplicate]

2.1 自定义month_diff函数

Define a "month" as 1/12 year, then do this:

def month_diff(d1, d2): 
    """Return the number of months between d1 and d2, 
    such that d2 + month_diff(d1, d2) == d1
    """
    diff = (12 * d1.year + d1.month) - (12 * d2.year + d2.month)
    return diff

assert diff_month(datetime(2010,10,1), datetime(2010,9,1)) == 1
assert diff_month(datetime(2010,10,1), datetime(2009,10,1)) == 12
assert diff_month(datetime(2010,10,1), datetime(2009,11,1)) == 11
assert diff_month(datetime(2010,10,1), datetime(2009,8,1)) == 14

date1 = datetime(2012, 2, 15)
date1.year # 2012, type为int

2.2 使用dateutile.relativedelta.relativedelta

from datetime import datetime
from dateutil import relativedelta
date1 = datetime.strptime(str('2011-08-15 12:00:00'), '%Y-%m-%d %H:%M:%S')
date2 = datetime.strptime(str('2012-02-15'), '%Y-%m-%d')
r = relativedelta.relativedelta(date2, date1)
print(r.months) # 5


date1 = datetime(2012, 2, 15)
date2 = datetime(2013, 12, 1)
r = relativedelta.relativedelta(date2, date1) # relativedelta(years=+1, months=+9, days=+16)
r.years * 12 + r.months # 21

有一个缺点是，如果时间差超过1年，则会将年的数字分配到r.years中，如2011-01和2012-01的month差r.months是0

3. datetime类型

In [1]: import datetime

In [2]: date1 = datetime.datetime(2010, 10, 4, 10, 10)

In [3]: date1
Out[3]: datetime.datetime(2010, 10, 4, 10, 10)

In [4]: date1.date()
Out[4]: datetime.date(2010, 10, 4)

In [6]: date1.day
Out[6]: 4

In [7]: date1.year
Out[7]: 2010

In [8]: type(date1.month)
Out[8]: int

4. 时间转换样例

A 目标

    Start Date  End Date    Contract
0   1/1/17  6/1/17  1
1   7/1/17  12/1/17 2

转换为：

    Month   Contract
0   2017-01 1
1   2017-02 1
2   2017-03 1
3   2017-04 1
4   2017-05 1
5   2017-06 1
6   2017-07 2
7   2017-08 2
8   2017-09 2
9   2017-10 2
10  2017-11 2
11  2017-12 2

B 代码

import pandas as pd
import numpy as np
import re, io, os

import datetime

data = '''Start Date    End Date    Contract
1/1/17  6/1/17  1
7/1/17  12/1/17 2
'''

df = pd.read_csv(io.StringIO(data), sep='\t', dtype=str)


for col in ['Start Date', 'End Date']:
    df[col] = df[col].map(lambda x: datetime.datetime.strptime(x, "%m/%d/%y"))

def convert_time(obj):
    x = pd.period_range(obj['Start Date'], obj['End Date'], freq='M')
    dfx = pd.DataFrame(x)
    dfx['Contract'] = obj['Contract']
    dfx.rename(columns={0:'Month'}, inplace=True)
    return dfx

for i in range(df.shape[0]):
    if i == 0:
        dfn = convert_time(df.iloc[i])
    else:
        dfn = pd.concat([dfn, convert_time(df.iloc[i])], ignore_index=True)
dfn