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I have a list of key:

list_date = ["MON", "TUE", "WED", "THU","FRI"]

I have many lists of values that created by codes below:

list_value = list()
for i in list(range(5, 70, 14)):
    list_value.append(list(range(i, i+10, 3)))

Rules created that:

  • first number is 5, a list contains 4 items has value equal x = x + 3, and so on [5, 8, 11, 14]

  • the first number of the second list equal: x = 5 + 14, and value inside still as above x = x +3

    [[5, 8, 11, 14], [19, 22, 25, 28], [33, 36, 39, 42], [47, 50, 53, 56], [61, 64, 67, 70]]

I expect to obtain a dict like this:

collections = {"MON":[5, 8, 11, 14], "TUE" :[19, 22, 25, 28], "WED":[33, 36, 39, 42], "THU":[47, 50, 53, 56], "FRI":[61, 64, 67, 70]}

Then, I used:

zip_iterator = zip(list_date, list_value)
collections = dict(zip_iterator)

To get my expected result.

I tried another way, like using the lambda function.

for i in list(range(5, 70, 14)):
    list_value.append(list(range(i,i+10,3)))
    couple_start_end[lambda x: x in list_date] = list(range(i, i + 10, 3))

And the output is:

{<function <lambda> at 0x000001BF7F0711F0>: [5, 8, 11, 14], <function <lambda> at 0x000001BF7F071310>: [19, 22, 25, 28], <function <lambda> at 0x000001BF7F071280>: [33, 36, 39, 42], <function <lambda> at 0x000001BF7F0710D0>: [47, 50, 53, 56], <function <lambda> at 0x000001BF7F0890D0>: [61, 64, 67, 70]}

I want to ask there is any better solution to create lists of values with the rules above? and create the dictionary collections without using the zip method?

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2 Answers 2

Reset to default
5
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One alternative approach is to take advantage of enumerate() -- which allows you to iterate over what could be thought of as a zip() of a list's indexes and values. To my eye, this version seems a bit clearer and simpler.

days = ['MON', 'TUE', 'WED', 'THU', 'FRI']

d = {
    day : [
        (5 + i * 14) + (j * 3)
        for j in range(4)
    ]
    for i, day in enumerate(days)
}

print(d)
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  • \$\begingroup\$ Personally, I always use itertools.product whenever there is a cartesian product involved: d = {day: (5 + i * 14) + (j * 3) for i,(j,day) in it.product(range(4), enumerate(days))}. \$\endgroup\$ Commented Apr 8, 2021 at 17:17
  • \$\begingroup\$ Just noticed the answers do not match. I retract my statement. \$\endgroup\$ Commented Apr 8, 2021 at 17:19
  • \$\begingroup\$ @Kevin: excuse me! but what is "it" in "it.product()" \$\endgroup\$ Commented Apr 8, 2021 at 23:45
  • \$\begingroup\$ @Scorpisces import itertools as it. But product() doesn't really help here. \$\endgroup\$ Commented Apr 9, 2021 at 1:07
  • 1
    \$\begingroup\$ Also you needn't write 3*j at all - you can instead move that to j in range(0, 12, 3) to avoid the multiplication. \$\endgroup\$ Commented Apr 10, 2021 at 4:14
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Another way that, as FMc's, has the advantage of not having to calculate/specify range limits (your 70 and i+10):

from itertools import count, islice

days = ['MON', 'TUE', 'WED', 'THU', 'FRI']

d = {day: list(islice(count(start, 3), 4))
     for day, start in zip(days, count(5, 14))}

print(d)
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