Как преобразовать два столбца из десятичных лет в дату

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Демонстрация.

0
задан Ole V.V. 27 February 2019 в 10:12
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2 ответа

Я положил свои объяснения в код ниже

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)
0
ответ дан Ilia Gilmijarow 27 February 2019 в 10:12
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Укажите правильный путь к вашему файлу и выходному файлу. Код ниже сделает все остальное.

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)
0
ответ дан Avinash Kumar 27 February 2019 в 10:12
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