Python中两个日期之间的差异

问题:Python中两个日期之间的差异

我有两个不同的日期,我想知道它们之间的天数差异。日期格式为YYYY-MM-DD。

我有一个可以在日期中添加或减去给定数字的功能:

def addonDays(a, x):
   ret = time.strftime("%Y-%m-%d",time.localtime(time.mktime(time.strptime(a,"%Y-%m-%d"))+x*3600*24+3600))      
   return ret

其中A是日期,x是我要添加的天数。结果是另一个日期。

我需要一个可以给出两个日期的函数,结果将是一个以天为单位的日期差的整数。

I have two different dates and I want to know the difference in days between them. The format of the date is YYYY-MM-DD.

I have a function that can ADD or SUBTRACT a given number to a date:

def addonDays(a, x):
   ret = time.strftime("%Y-%m-%d",time.localtime(time.mktime(time.strptime(a,"%Y-%m-%d"))+x*3600*24+3600))      
   return ret

where A is the date and x the number of days I want to add. And the result is another date.

I need a function where I can give two dates and the result would be an int with date difference in days.


回答 0

使用-得到两者之间的区别datetime对象,并采取days会员。

from datetime import datetime

def days_between(d1, d2):
    d1 = datetime.strptime(d1, "%Y-%m-%d")
    d2 = datetime.strptime(d2, "%Y-%m-%d")
    return abs((d2 - d1).days)

Use - to get the difference between two datetime objects and take the days member.

from datetime import datetime

def days_between(d1, d2):
    d1 = datetime.strptime(d1, "%Y-%m-%d")
    d2 = datetime.strptime(d2, "%Y-%m-%d")
    return abs((d2 - d1).days)

回答 1

另一个简短的解决方案:

from datetime import date

def diff_dates(date1, date2):
    return abs(date2-date1).days

def main():
    d1 = date(2013,1,1)
    d2 = date(2013,9,13)
    result1 = diff_dates(d2, d1)
    print '{} days between {} and {}'.format(result1, d1, d2)
    print ("Happy programmer's day!")

main()

Another short solution:

from datetime import date

def diff_dates(date1, date2):
    return abs(date2-date1).days

def main():
    d1 = date(2013,1,1)
    d2 = date(2013,9,13)
    result1 = diff_dates(d2, d1)
    print '{} days between {} and {}'.format(result1, d1, d2)
    print ("Happy programmer's day!")

main()

回答 2

我尝试了上面larsmans发布的代码,但是有两个问题:

1)原样的代码将引发mauguerra提到的错误2)如果将代码更改为以下内容:

...
    d1 = d1.strftime("%Y-%m-%d")
    d2 = d2.strftime("%Y-%m-%d")
    return abs((d2 - d1).days)

这会将您的datetime对象转换为字符串,但是有两件事

1)尝试执行d2-d1将失败,因为您无法在字符串上使用减号运算符,并且2)如果阅读了上述答案的第一行,则想在两个datetime对象上使用-运算符,但是将它们转换为字符串

我发现您实际上只需要以下内容:

import datetime

end_date = datetime.datetime.utcnow()
start_date = end_date - datetime.timedelta(days=8)
difference_in_days = abs((end_date - start_date).days)

print difference_in_days

I tried the code posted by larsmans above but, there are a couple of problems:

1) The code as is will throw the error as mentioned by mauguerra 2) If you change the code to the following:

...
    d1 = d1.strftime("%Y-%m-%d")
    d2 = d2.strftime("%Y-%m-%d")
    return abs((d2 - d1).days)

This will convert your datetime objects to strings but, two things

1) Trying to do d2 – d1 will fail as you cannot use the minus operator on strings and 2) If you read the first line of the above answer it stated, you want to use the – operator on two datetime objects but, you just converted them to strings

What I found is that you literally only need the following:

import datetime

end_date = datetime.datetime.utcnow()
start_date = end_date - datetime.timedelta(days=8)
difference_in_days = abs((end_date - start_date).days)

print difference_in_days

回答 3

试试这个:

data=pd.read_csv('C:\Users\Desktop\Data Exploration.csv')
data.head(5)
first=data['1st Gift']
last=data['Last Gift']
maxi=data['Largest Gift']
l_1=np.mean(first)-3*np.std(first)
u_1=np.mean(first)+3*np.std(first)


m=np.abs(data['1st Gift']-np.mean(data['1st Gift']))>3*np.std(data['1st Gift'])
pd.value_counts(m)
l=first[m]
data.loc[:,'1st Gift'][m==True]=np.mean(data['1st Gift'])+3*np.std(data['1st Gift'])
data['1st Gift'].head()




m=np.abs(data['Last Gift']-np.mean(data['Last Gift']))>3*np.std(data['Last Gift'])
pd.value_counts(m)
l=last[m]
data.loc[:,'Last Gift'][m==True]=np.mean(data['Last Gift'])+3*np.std(data['Last Gift'])
data['Last Gift'].head()

Try this:

data=pd.read_csv('C:\Users\Desktop\Data Exploration.csv')
data.head(5)
first=data['1st Gift']
last=data['Last Gift']
maxi=data['Largest Gift']
l_1=np.mean(first)-3*np.std(first)
u_1=np.mean(first)+3*np.std(first)


m=np.abs(data['1st Gift']-np.mean(data['1st Gift']))>3*np.std(data['1st Gift'])
pd.value_counts(m)
l=first[m]
data.loc[:,'1st Gift'][m==True]=np.mean(data['1st Gift'])+3*np.std(data['1st Gift'])
data['1st Gift'].head()




m=np.abs(data['Last Gift']-np.mean(data['Last Gift']))>3*np.std(data['Last Gift'])
pd.value_counts(m)
l=last[m]
data.loc[:,'Last Gift'][m==True]=np.mean(data['Last Gift'])+3*np.std(data['Last Gift'])
data['Last Gift'].head()

回答 4

pd.date_range(’2019-01-01’,’2019-02-01’)。shape [0]

pd.date_range(‘2019-01-01’, ‘2019-02-01’).shape[0]