在Python中使用函数比较时间戳（与Delorean）

Question

我有一段基本的Python代码，我试图将一段时间与另一段时间进行比较（我必须承认我对Python有点陌生）。我使用Delorean将字符串转换为epoch整数，并将其与变量lastTime进行比较。我的代码的其余部分可以工作，但只要我尝试添加新函数，我的读数就会变成空白。我正在努力解决这个问题，因为Python没有抛出任何错误，它只是什么也没有给我。你知道吗

下面是它的代码：

import random,sys,csv, delorean
from collections import defaultdict
from delorean import Delorean
from delorean import parse

size = ['small','medium','large']
color = ['blue','red','green']
body = ['fish','squid']

fishparts = defaultdict(set)
lastfish = defaultdict(str)
lastTime = 0


def tenMinInterval(ts, lt): # HAVING PROBLEMS HERE
    global lastTime
    curTime = Delorean(ts).epoch()
    if curTime > lt+600000: # timestamp is a string and lastTime is an int
        # return True
        lastTime = curTime
        return 'Upadted Time'
    else:
        # return False
        return 'Not Upadted'

def complexityFish(ps,pf):
    score = 1
    if ps == 'medium':
        score += 1
    elif ps == 'large':
        score += 2
    if pf == 'squid':
        score += 2
    return str(score)

def diffPrev(a,la,b,lb,c,lc):
    score = 0
    if a != la:
        score += 1
    if b != lb:
        score += 1
    if c != lc:
        score += 1
    return str(score)

def diffUniq(player,x):
    score = 0
    for e in x:
        if e not in fishparts[player]:
            score += 1
        fishparts[player].add(e)
    return str(score)


def parseOneFish(p_player,p_fish):
    player  = p_player
    fish    = p_fish
    if lastfish[player] != '':
        ls,lc,lt = lastfish[player].split(' ')
    else:
        ls = lc = lt = ''
    s,c,t = fish.split(' ')
    lastfish[player] = fish
    return((complexityFish(s,t),diffPrev(s,ls,c,lc,t,lt),diffUniq(player,[s,c,t])))


csvfilename = sys.argv[1]
csvdata = csv.DictReader(open(csvfilename,'rb'),delimiter=',')
x = False
for line in csvdata:
    if not x:
        print ','.join([k for k in line]),
        print ',complexity,diffprev,diffuniq'
        x = True
    try:
        cx,dp,du = parseOneFish(line['playerID'],line['fishType'])
        tm = tenMinInterval(line['timestamp'], lastTime) # HAVING PROBLEMS HERE
        print ','.join([line[k] for k in line]) + ',',
        print ','.join([cx,dp,du])
        print ','.join(tm)
    except:
        print ''  

我正在努力解决的部分在函数tenMinInterval和底部的tm = tenMinInterval(line['timestamp'], lastTime) 我知道现在回报率不高，但这不应该影响其他的正确吗？你知道吗

这是我的csv file as well的一个示例

Answer 1

因此，多亏了mkrieger1，我能够更好地检查我的错误，结果是我的DB如何构造日期以及Delorian default如何读取日期的解析问题。你知道吗

下面是修复的函数：

def tenMinInterval(ts, lt):
    global lastTime
    strTime = delorean.interface.parse(ts, dayfirst=False, yearfirst=False)
    curTime = strTime.epoch()
    print curTime
    if curTime > lt+600: # timestamp is a string and lastTime is an int
        # return True
        lastTime = curTime
        print 'Time Updated'
    else:
        # return False
        print 'Time not Updated'

如果有人需要它，有一个解析日期here和here的解释。在修补时，我还需要添加from delorean import epoch