怎样用spss中的python语法将V2列的时间戳转为时间（年月日时分秒）

Question

怎样用spss中的python语法将V2列的时间戳转为时间（年月日时分秒）
用spss中新建语法的方式

Answer 1

# -*- coding: utf-8 -*-
import math

#实现整除运算
def div(x,y):
return int(round(x)/round(y))
#return $cal2('x','x',p_div,x,y);
def DF2DHMS(F):
df = F
day = math.floor(df)
hour = math.floor((df-day)*24)
minute = math.floor((df-day-hour/24)*1440)
sec = (df-day-hour/24-minute/1440)*86400
return [day,hour,minute,sec]
MJD=17366.62152773142
DJMIN = -68569.5
DJMAX = 1e9

DJ1 = 2400000.5
DJ2 = MJD
DJ = DJ1 + DJ2
D1 =''
D2 =''
J =''
JD =''
if ( DJ < DJMIN or DJ > DJMAX ):

J = -1
print u'无效的日期: '+MJD
print J
else:
J = 0
if ( DJ1 >= DJ2 ):
D1 = DJ1
D2 = DJ2
else:
D1 = DJ2
D2 = DJ1
D2 = D2 - 0.5
F1 = D1%1.0
F2 = D2%1.0
F = (F1+F2) % 1.0
if ( F < 0 ) :F = F + 1.0
D = round(D1-F1) + round(D2-F2) + round(F1+F2-F)
JD = round(D) + 1
L = JD + 68569
N = div( 4*L , 146097)
L = L - div(( 146097*N + 3 ) , 4)
I = div( 4000 * (L+1) , 1461001)
L = L - div( 1461*I , 4) + 31
K = div( 80*L , 2447)
ID = L - div( 2447*K , 80)
L = div(K , 11)

IM = K + 2 - 12*L
IY = 100 * ( N-49 ) + I + L
FD = DF2DHMS(F)
print MJD,'对应日期为',[IY,IM,int(ID),int(FD[1]),int(FD[2]),FD[3]]

-------

>>>
17366.6215277 对应日期为 [1906, 6, 5, 14, 54, 59.99599456786802]
>>>
http://www.scicalweb.com/html/online-calculate/609.html

Answer 2

格式问题，可以调整的本回答被网友采纳