Calculating Time Difference

at the start and end of my program, I have

from time import strftime
print int(strftime("%Y-%m-%d %H:%M:%S")



Y1=int(strftime("%Y"))
m1=int(strftime("%m"))
d1=int(strftime("%d"))
H1=int(strftime("%H"))
M1=int(strftime("%M"))
S1=int(strftime("%S"))


Y2=int(strftime("%Y"))
m2=int(strftime("%m"))
d2=int(strftime("%d"))
H2=int(strftime("%H"))
M2=int(strftime("%M"))
S2=int(strftime("%S"))

print "Difference is:"+str(Y2-Y1)+":"+str(m2-m1)+":"+str(d2-d1)
          +" "+str(H2-H1)+":"+str(M2-M1)+":"+str(S2-S1)

But when I tried to get the difference, I get syntax errors…. I am doing a few things wrong, but I’m not sure what is going on…

Basically, I just want to store a time in a variable at the start of my program, then store a 2nd time in a second variable near the end, then at the last bit of the program, compute the difference and display it. I am not trying to time a function speed. I am trying to log how long it took for a user to progress through some menus. What is the best way to do this?

Answers:

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Method 1

The datetime module will do all the work for you:

>>> import datetime
>>> a = datetime.datetime.now()
>>> # ...wait a while...
>>> b = datetime.datetime.now()
>>> print(b-a)
0:03:43.984000

If you don’t want to display the microseconds, just use (as gnibbler suggested):

>>> a = datetime.datetime.now().replace(microsecond=0)
>>> b = datetime.datetime.now().replace(microsecond=0)
>>> print(b-a)
0:03:43

Method 2

from time import time

start_time = time()
...

end_time = time()
seconds_elapsed = end_time - start_time

hours, rest = divmod(seconds_elapsed, 3600)
minutes, seconds = divmod(rest, 60)

Method 3

You cannot calculate the differences separately … what difference would that yield for 7:59 and 8:00 o’clock? Try

import time
time.time()

which gives you the seconds since the start of the epoch.

You can then get the intermediate time with something like

timestamp1 = time.time()
# Your code here
timestamp2 = time.time()
print "This took %.2f seconds" % (timestamp2 - timestamp1)

Method 4

Both time.monotonic() and time.monotonic_ns() are correct. Correct as in monotonic.

>>> import time
>>>
>>> time.monotonic()
452782.067158593
>>>
>>> t0 = time.monotonic()
>>> time.sleep(1)
>>> t1 = time.monotonic()
>>> print(t1 - t0)
1.001658110995777

Regardless of language, monotonic time is the right answer, and real time is the wrong answer. The difference is that monotonic time is supposed to give a consistent answer when measuring durations, while real time isn’t, as real time may be adjusted – indeed needs to be adjusted – to keep up with reality. Monotonic time is usually the computer’s uptime.

As such, time.time() and datetime.now() are wrong ways to do this.

Python also has time.perf_counter() and time.perf_counter_ns(), which are specified to have the highest available resolution, but aren’t guarranteed to be monotonic. On PC hardware, though, both typically have nanosecond resolution.

Method 5

Here is a piece of code to do so:

def(StringChallenge(str1)):

#str1 = str1[1:-1]
h1 = 0
h2 = 0
m1 = 0
m2 = 0

def time_dif(h1,m1,h2,m2):
    if(h1 == h2):
        return m2-m1
    else:
        return ((h2-h1-1)*60 + (60-m1) + m2)
count_min = 0

if str1[1] == ':':
    h1=int(str1[:1])
    m1=int(str1[2:4])
else:
    h1=int(str1[:2])
    m1=int(str1[3:5])

if str1[-7] == '-':
    h2=int(str1[-6])
    m2=int(str1[-4:-2])
else:
    h2=int(str1[-7:-5])
    m2=int(str1[-4:-2])

if h1 == 12:
    h1 = 0
if h2 == 12:
    h2 = 0

if "am" in str1[:8]:
    flag1 = 0
else:
    flag1= 1

if "am" in str1[7:]:
    flag2 = 0
else:
    flag2 = 1

if flag1 == flag2:
    if h2 > h1 or (h2 == h1 and m2 >= m1):
        count_min += time_dif(h1,m1,h2,m2)
    else:
        count_min += 1440 - time_dif(h2,m2,h1,m1)
else:
    count_min += (12-h1-1)*60
    count_min += (60 - m1)
    count_min += (h2*60)+m2


return count_min

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0