如何将两列从十进制年份转换为日期答案

【问题标题】：How to convert two columns from decimal years to date如何将两列从十进制年份转换为日期
【发布时间】：2019-07-20 01:50:09
【问题描述】：

我是 Python 新手，但遇到了问题。

我在 .txt 文档中有两列十进制年份的数据，我想将两列中的每个数字转换为数据 (yyyy-mm-dd)

2014.16020 2019.07190
2000.05750 2019.10750
2001.82610 2019.10750
2010.36280 2019.07190
2005.24570 2019.10750
2015.92610 2019.10750
2003.43600 2014.37100

然后从第一列的数据中减去第二列的数据，得到两个数据之间的天数。例如结果应该是这样的：

【问题讨论】：

什么是小数年？我不明白它是如何转换为日期的。

标签： python datetime converters date-difference

【解决方案1】：

提供文件和输出文件的正确路径。下面的代码将完成其余的工作。

import pandas as pd
import numpy as np
df=pd.read_csv('your_file.txt',delimiter=' ',header=None,parse_dates=[0,1])
df['date_diffrence']=((df[1]-df[0])/np.timedelta64(1,'D')).astype(int)
df.to_csv('your_file_result.txt',header=None,sep=' ',index=False)

【讨论】：

【解决方案2】：

我把我的解释放在下面的代码中

from datetime import datetime  # yes they named a class the same as module
x = '''2014.16020 2019.07190
2000.05750 2019.10750
2001.82610 2019.10750
2010.36280 2019.07190
2005.24570 2019.10750
2015.92610 2019.10750
2003.43600 2014.37100'''

# split input into lines. Assumption here is that there is one pair of dates per each line
lines = x.splitlines() 
# set up a container (list) for outputs
deltas = []
# process line by line
for line in lines:
    # split line into separate dates
    inputs = line.split()
    dates = []
    for input in inputs:
        # convert text to number
        date_decimal = float(input)
        # year is the integer part of the input
        date_year = int(date_decimal)
        # number of days is part of the year, which is left after we subtract year
        year_fraction = date_decimal - date_year
        # a little oversimplified here with int and assuming all years have 365 days
        days = int(year_fraction * 365)
        # now convert the year and days into string and then into date (there is probably a better way to do this - without the string step)
        date = datetime.strptime("{}-{}".format(date_year, days),"%Y-%j")  
        # see https://docs.python.org/3/library/datetime.html#strftime-strptime-behavior for format explanation
        dates.append(date)
    deltas.append(dates[1] - dates[0])

# now print outputs
for delta in deltas:
    print(delta.days)

【讨论】：