CODEWITHSUNDEEP
pythonlistloopsfor-loopappend
Wayne
Wayne

Reputation: 43

Python two Lists In two For Loops

I have two list here:

list1 = [u'2018-05-06T15:53:05.000-0400', '2018-05-06T17:53:05.000-0400']

list2 = [u'2018-05-06T15:32:24.000-0400', u'2018-05-06T15:32:29.000-0400', u'2018-05-06T15:32:36.000-0400', u'2018-05-06T15:53:05.000-0400', u'2018-05-06T16:42:41.000-0400', u'2018-05-10T00:16:12.000-0400', u'2018-05-10T00:16:23.000-0400', u'2018-05-12T17:37:42.000-0400']

I'm trying to compare the elements in two lists and return the minimum value of the different for both element 0 in list1 and element 1 in list2, for example:

for value1 in list1:
    value1 = datetime.strptime(value1, '%Y-%m-%dT%H:%M:%S.000-0400')
    for value2 in list2:
        value2 = datetime.strptime(value2, '%Y-%m-%dT%H:%M:%S.000-0400')
        if value1 < value2:

            value3 = value2 - value1
            list3.append(value3)
 print list3

print list3 give me:

[datetime.timedelta(0, 2976), datetime.timedelta(3, 30187), datetime.timedelta(3, 30198), datetime.timedelta(6, 6277), datetime.timedelta(3, 22987), datetime.timedelta(3, 22998), datetime.timedelta(5, 85477)]

Everything in the same list. Here in list3[0], I could get the minimum value for first element in list1 but how could I get the minimum value for the second one list3[3] and loop if i have element 3 in list1?

I would like to separate the list for compare result for first element in list1 and second element, please let me know how to achieve this, much appreciated!

Upvotes: 2

Views: 1124

Answers (3)

rafaelc
rafaelc

Reputation: 59274

If I understand correctly, you can separate your time deltas in different lists, and then just use min to get the minimum values.

Using your own code:

list3=[]
for value1 in list1:
    value1 = datetime.strptime(value1, '%Y-%m-%dT%H:%M:%S.000-0400')
    aux_list = []
    for value2 in list2:
        value2 = datetime.strptime(value2, '%Y-%m-%dT%H:%M:%S.000-0400')
        if value1 < value2:

            value3 = value2 - value1
            aux_list.append(value3)
    list3.append(aux_list)

This way, your list3 holds the difference from each element of list1 to all elements of list2

>>> print(list3)
[[datetime.timedelta(0, 2976), datetime.timedelta(3, 30187), datetime.timedelta(3, 30198), datetime.timedelta(6, 6277)], 
 [datetime.timedelta(3, 22987), datetime.timedelta(3, 22998), datetime.timedelta(5, 85477)]]

Then you can get the minimum difference from the first item using min(list[0]) and for the second min(list[1])

Notice that, if you have

l1 = [datetime.strptime(x, '%Y-%m-%dT%H:%M:%S.000-0400') for x in list1]
l2 = [datetime.strptime(x, '%Y-%m-%dT%H:%M:%S.000-0400') for x in list2]

your whole code can be written in one line using list comprehension:

k = [min([v2 - v1 for v2 in l2 if v2>v1]) for v1 in l1]

being l1 and l2 your list of datetime objects. This yields

0:49:36 # k[0]
3 days, 6:23:07 # k[1]

Upvotes: 2

mentoc3000
mentoc3000

Reputation: 334

If you want to iterate over the two lists together you can zip them and iterate over them simultaneously. Then compare the values and append the one you want to the result list.

list1 = [1, 2, 3]
list2 = [0, 3]
list3 = []
for (x, y) in zip(list1, list2):
    z = min(x, y) 
    list3.append(z)

Note that the result will be the size of the smaller list, list2 in this case.

Upvotes: 0

AmphotericLewisAcid
AmphotericLewisAcid

Reputation: 2299

First off, Python doesn't know that the string represents a date. You can't just throw data into a string and expect the interpreter to know what you want. So, you're just going to get an error because the "-" operator means nothing to strings.

If you want to compare two dates, I suggest reading up on Python's builtin datetime module: https://docs.python.org/2/library/datetime.html

Upvotes: 0

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