How to break loop if variable changes

Question

I am running a for loop, in which the iterator loops through list of time based values.

for example this is what my list looks like

lst = ['00:00:01', '00:00:05', '00:00:07', '00:01:10', ... , '00:20:23']

I want to check if the minute part of the time data changes. If it changes then break the loop.

this is my code so far

time_lst = []

for time in lst:
    minute = time[3:5]

    if minute changes:
        time_lst.append(time)
        break

print(time_lst)

in the above program, how do I program the if condition?

I looked online but I could not find a good solution to this problem.

assign a value to "minute" before the loop starts. during loop, check the "current" minute against the "previous" assigned minute. If they match, proceed. If they mismatch, break. — Paritosh Singh
– Paritosh Singh, Commented Aug 9, 2019 at 8:38
Essentially, you need a variable that keeps "state" of the minute value during iteration — Paritosh Singh
– Paritosh Singh, Commented Aug 9, 2019 at 8:38
@Sashaank Please, mark an answer, if your issue is resolved. — Siddharth Dinesh
– Siddharth Dinesh, Commented Aug 9, 2019 at 8:53
Only not sure if you want keep tracking of times before change or after change as in ['00:00:07', '00:00:50' , '00:01:10'] if you want to get 00:00:50 or 00:01:10 in list @Sashaank — Marek Maszay
– Marek Maszay, Commented Aug 9, 2019 at 9:19

Ardis thrash · Accepted Answer · 2019-08-09 08:43:17Z

4

This should work.

for time in lst:
    minute = time[3:5]
    if minute != lst[0][3:5]:
        break
    print minute

answered Aug 9, 2019 at 8:43

Ardis thrash

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1 Comment

Siddharth Dinesh Over a year ago

The only answer that, gives the solution to the question and it's from a new contributor

Marek Maszay · Accepted Answer · 2019-08-09 08:57:13Z

2

EDIT

With edited question so that it keeps changes of time between minutes

time_lst = []
lastMinute = lst[0][3:5] #so that it won't break on first item
for time in lst:
    minute = time[3:5]

    if minute != lastMinute: # `<>` still usable in python2, `!=` is better for compatibility between python2 and python3
        time_lst.append(time)
    lastMinute = minute

print(time_lst)

edited Aug 9, 2019 at 8:57

answered Aug 9, 2019 at 8:39

Marek Maszay

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3 Comments

WiseDev Over a year ago

do you mean != ?

Marek Maszay Over a year ago

@BlueRineS <> can be used still in 2.7 but yeah for python3+ it should be !=

WiseDev Over a year ago

Great. Maybe you should specify that in your answer

Amin Rashidbeigi · Accepted Answer · 2019-08-09 08:56:20Z

2

You can simply do it by saving previous minute and compare it with new value:

previous_minute = lst[0][3:5]
for time in lst:
    minute = time[3:5]
    if previous_minute not minute:
        break
    previous_minute = minute

If you want to save changed times:

changed_times = []
previous_minute = lst[0][3:5]
for time in lst:
    minute = time[3:5]
    if previous_minute not minute:
        changed_times.append(time)
    previous_minute = minute

edited Aug 9, 2019 at 8:56

answered Aug 9, 2019 at 8:39

Amin Rashidbeigi

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Comments

Ahmad Farhan · Accepted Answer · 2019-08-09 08:42:23Z

1

Save the first element's minute value before starting the loop. Compare all following values with it. break if it doesn't match.

prev=lst[0][3:5]
for time in lst:
    if(time[3:5]!=prev):
        break

answered Aug 9, 2019 at 8:42

Ahmad Farhan

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Comments

Neo · Accepted Answer · 2019-08-09 08:51:53Z

You could keep a variable out of the loop, holding the first minute. Then the test is easy - you compare the new minute to the first one. Once they're inequal, it means the minute changed.

first_minute = lst[0][3:5]
for time in lst:
    minute = time[3:5]

    if minute != first_minute:
        break

However, if all you're doing in the loop is updating another list, you could use list comprehension to filter the non-minute-equal times:

new_list = [time in lst if time[3:5] == lst[0][3:5]]

You could also save lst[0][3:5] to a first_minute variable as I've done above. Another modification is to use filter and lambda t: t[3:5] == lst[0][3:5].

NOTICE: The list comprehension solution filters all the list. It means if after the minute change, if the times go back to the same minute value (maybe with a different hour value?), it will include them as well - while the loop & break solution won't. Use according to your needs.

To clarify, using the first solution with lst = ["00:00:00", "00:00:04", "00:01:00", "01:00:02"] will result in iterating over only ["00:00:00", "00:00:04"], while the second will give (for the same lst) ["00:00:00", "00:00:04", "01:00:02"].

alec_djinn · Accepted Answer · 2019-08-09 10:14:30Z

0

You need to keep track of the last value.

There are several ways to accomplish this. The most straight forward is to update a variable.

last_value = False
for time in lst:
    minute = time[3:5]
    if not last_value:
        last_value = minute
    else:
        if minute != last_value:
            ...
            break

edited Aug 9, 2019 at 10:14

answered Aug 9, 2019 at 8:41

alec_djinn

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