Accessing index using list comprehesion in Python

Question

Hello recently I am trying to figure out list comprehension. And it looks like I suck at this :/

Here is my code I am trying to remake using list comprehension

base = datetime.datetime.today()
date_list = [base - datetime.timedelta(days=x) for x in range(61)]
del date_list[0]
date_list.reverse()

weekend = []
dates = []
for idx, val in enumerate(date_list):
    dates.append(str(val)[0:10])
    dates.append(str(val)[0:10])
    case = val.isoweekday()
    if case == 6 or case == 7:
        weekend.append(str(val)[0:10])

I looked it out how to use enumerate using this method and i found this:

[val for idx, val in enumerate(date_list)]

But i don't know how to continue this idea.

I would really appreciate some help :)

Why the use of enumerate? doesn't look like you need the index idx. — ChrisOram
– ChrisOram, Commented Aug 13, 2021 at 8:05
Well.. you are right. This is a leftover from previous version. I didn't really think about this. Thanks — antekkalafior
– antekkalafior, Commented Aug 13, 2021 at 8:06
Notwithstanding the fact that you don't need to use enumerate, why use a list comprehension when your loop does what you need? Sure, it can be done using list comprehension but it'll end up being unreadable/unmaintainable — user2668284
– user2668284, Commented Aug 13, 2021 at 8:08

Daweo · Accepted Answer · 2021-08-13 08:20:47Z

This

dates = []
for idx, val in enumerate(date_list):
    dates.append(str(val)[0:10])
    dates.append(str(val)[0:10])

can be reworked into list comprehension but beware that you might end with low readibility. Consider similar but simpler case:

digits = [1,2,3,4,5,6,7,8,9]
numbers = []
for val in digits:
    numbers.append(val)
    numbers.append(val)
print(numbers)

output

[1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9]

First step is to rework code such that is there is exactly one .append, in this case replace for loop with

for val in digits:
    for _ in range(2):
        numbers.append(val)

As you might check output is same, final step is to rewrite that into list comprehension which in case of nested for is shallow-deep to left-right thus outcome is

digits = [1,2,3,4,5,6,7,8,9]
numbers = [val for val in digits for _ in range(2)]
print(numbers)

This approach might be also used if you have more levels of nesting.

JuR · Accepted Answer · 2021-08-13 12:39:09Z

1

For list comprehension with if condition it should look like this [function(x) if condition(x) else other_function(x) for x in list] So for your case, I would say:

base = datetime.datetime.today()
date_list = [base - datetime.timedelta(days=x) for x in range(1,61)]
date_list.reverse()

weekend = [date.strftime("%m/%d/%Y, %H:%M:%S")  for date in date_list if (date.weekday()==5 or date.weekday()==6)]
dates=[]
for date in date_list:
    dates.extend([date.strftime("%m/%d/%Y, %H:%M:%S"),date.strftime("%m/%d/%Y, %H:%M:%S")])

It's better to use strftime to transform datetime in the format you want: https://docs.python.org/fr/3.6/library/datetime.html#strftime-and-strptime-behavior

edited Aug 13, 2021 at 12:39

answered Aug 13, 2021 at 8:08

JuR

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4 Comments

antekkalafior Over a year ago

I copied ur code but it says there is a syntax error somwhere near for in date.weekday()==7) for date in date_list

Károly Szabó Over a year ago

The list comprehension is incorrect. This should be the right one: weekend = [date.strftime("%m/%d/%Y, %H:%M:%S") for date in date_list if (date.weekday()==6 or date.weekday()==7) ] The order of the for and if statement matters.

JuR Over a year ago

Indeed, my bad, to use this you absolutly need a else statement which is not really suitable in your case

antekkalafior Over a year ago

I also belive that .weekday() returns different values than .isoweekday(). Instead of (date.weekday()==6 or date.weekday()==7) 6 and 7 should be changed like so (date.weekday()==5 or date.weekday()==6)

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