i am trying to compute time intervals per day from a list of unix timestamps in Python. I have searched for simular questions on stack overflow but mostly found examples of computing deltas or SQL solutions.
I have a list of the sort:
timestamps = [1176239419.0, 1176334733.0, 1176445137.0, 1177619954.0, 1177620812.0, 1177621082.0, 1177838576.0, 1178349385.0, 1178401697.0, 1178437886.0, 1178926650.0, 1178982127.0, 1179130340.0, 1179263733.0, 1179264930.0, 1179574273.0, 1179671730.0, 1180549056.0, 1180763342.0, 1181386289.0, 1181990860.0, 1182979573.0, 1183326862.0]
I can easily turn this list of timestamps into datetime objects using:
[dt.datetime.fromtimestamp(int(i)) for i in timestamps]
From there I can probably write quite a lengthy code where the first day/month is kept and a check is done to see if the next item in the list is of the same day/month. If it is I look at the times, get the first and last from the day and store the interval + day/month in a dictionary.
As I am fairly new to Python I was wondering what is the best way to do this in this programming language without the abusive use of if/else statements.
Thank you in advance
如果列表排序为你的情况，那么你可以使用itertools.groupby()到组的时间戳到天：
#!/usr/bin/env python
from datetime import date, timedelta
from itertools import groupby
epoch = date(1970, 1, 1)
result = {}
assert timestamps == sorted(timestamps)
for day, group in groupby(timestamps, key=lambda ts: ts // 86400):
# store the interval + day/month in a dictionary.
same_day = list(group)
assert max(same_day) == same_day[-1] and min(same_day) == same_day[0]
result[epoch + timedelta(day)] = same_day[0], same_day[-1]
print(result)
产量
{datetime.date(2007, 4, 10): (1176239419.0, 1176239419.0),
datetime.date(2007, 4, 11): (1176334733.0, 1176334733.0),
datetime.date(2007, 4, 13): (1176445137.0, 1176445137.0),
datetime.date(2007, 4, 26): (1177619954.0, 1177621082.0),
datetime.date(2007, 4, 29): (1177838576.0, 1177838576.0),
datetime.date(2007, 5, 5): (1178349385.0, 1178401697.0),
datetime.date(2007, 5, 6): (1178437886.0, 1178437886.0),
datetime.date(2007, 5, 11): (1178926650.0, 1178926650.0),
datetime.date(2007, 5, 12): (1178982127.0, 1178982127.0),
datetime.date(2007, 5, 14): (1179130340.0, 1179130340.0),
datetime.date(2007, 5, 15): (1179263733.0, 1179264930.0),
datetime.date(2007, 5, 19): (1179574273.0, 1179574273.0),
datetime.date(2007, 5, 20): (1179671730.0, 1179671730.0),
datetime.date(2007, 5, 30): (1180549056.0, 1180549056.0),
datetime.date(2007, 6, 2): (1180763342.0, 1180763342.0),
datetime.date(2007, 6, 9): (1181386289.0, 1181386289.0),
datetime.date(2007, 6, 16): (1181990860.0, 1181990860.0),
datetime.date(2007, 6, 27): (1182979573.0, 1182979573.0),
datetime.date(2007, 7, 1): (1183326862.0, 1183326862.0)}
如果只有一个在那一天比它重复两次时间戳。
你会怎么做之后，以测试如果最后（例如）5个结果中的条目比以前的14间隔较大？
entries = sorted(result.items())
intervals = [(end - start) for _, (start, end) in entries]
print(max(intervals[-5:]) > max(intervals[-5-14:-5]))
# -> False
您可以使用collections.defaultdict 。 这是当你试图建立一个收集无大小和成员inital估计惊人地方便。
from collections import defaultdict
# Initialize default dict by the type list
# Accessing a member that doesn't exist introduces that entry with the deafult value for that type
# Here, when accessing a non-existant member adds an empty list to the collection
intervalsByDate = defaultdict(list)
for t in timestamps:
dt = dt.datetime.fromtimestamp(t)
myDateKey = (dt.day, dt.month, dt.year)
# If the key doesn't exist, a new empty list is added
intervalsByDate[myDateKey].append(t)
由此看来， intervalsByDate现在是一个dict与价值观作为基于日历日期排序的列表时间戳。 对于每一个日期，你可以时间戳进行排序，并得到总的时间间隔。 迭代defaultdict等同于dict （该子类的dict或多个）。
output = {}
for date, timestamps in intervalsByDate.iteritems():
sortedIntervals = sorted(timestamps)
output[date] = sortedIntervals[-1] - sortedIntervals[0]
现在output是地图与以毫秒为单位的值间隔日期。 用它做什么，你会的！
编辑
这是正常的键不排序？ 我从togheter混合不同月份的密钥。
是的，因为（哈希）地图和dicts 本质上是无序的
我怎么能够，例如，从一个月选择第一个24天，然后最后
如果我是对我的回答很坚决，我想也许看看这个，这是一个有序的默认字典。 。 然而，你可以修改的数据类型output到一些东西，是不是dict ，以满足您的需求。 例如，一个list ，并责令其基于日期。
只是减去对方2个日期。 这将导致timedelta实例。 见datetime.timedelta： https://docs.python.org/2/library/datetime.html#timedelta-objects
from datetime import datetime
delta = datetime.today() - datetime(year=2015, month=01, day=01)
#Actual printed out values may change depending o when you execute this :-)
print delta.days, delta.seconds, delta.microseconds #prints 49 50817 381000
print delta.total_seconds() #prints 4284417.381 which is 49*24*3600 + 50817 + 381000/1000000
具有行切片和zip结合这让您的解决方案。 一个例子的解决办法是：
timestamps = [1176239419.0, 1176334733.0, 1176445137.0, 1177619954.0, 1177620812.0, 1177621082.0, 1177838576.0, 1178349385.0, 1178401697.0, 1178437886.0, 1178926650.0, 1178982127.0, 1179130340.0, 1179263733.0, 1179264930.0, 1179574273.0, 1179671730.0, 1180549056.0, 1180763342.0, 1181386289.0, 1181990860.0, 1182979573.0, 1183326862.0]
timestamps_as_dates = [datetime.fromtimestamp(int(i)) for i in timestamps]
# Make couples of each timestamp with the next one
# timestamps_as_dates[:-1] -> all your timestamps but the last one
# timestamps_as_dates[1:] -> all your timestamps but the first one
# zip them together so that first and second are one couple, then second and third, ...
intervals = zip(timestamps_as_dates[:-1],timestamps_as_dates[1:])
interval_timedeltas = [(interval[1]-interval[0]).total_seconds() for interval in intervals]
# result = [95314.0, 110404.0, 1174817.0, 858.0, 270.0, 217494.0, 510809.0, 52312.0, 36189.0, 488764.0, 55477.0, 148213.0, 133393.0, 1197.0, 309343.0, 97457.0, 877326.0, 214286.0, 622947.0, 604571.0, 988713.0, 347289.0]
这也适用于添加或减去一个日期一定时期：
from datetime import datetime, timedelta
tomorrow = datetime.today() + timedelta(days=1)
我没有增加或减去月或数年的简单的解决方案。
