Get list from dictionary with list

Question

My data is organized like this:

d={1: [44, 56.5238], 2: [63, 56.8663], 3: [93, 55.7313], 4: [112, 55.5425]}

I want to create list with first values in these small lists:

new_list=[44, 63, 93, 112]

My ugly way to do it:

new_list=[] for item in d.values(): new_list.append(item[0])

Can I do it in better and shorter way, may be in one line?

With a list comprehension. However, what version of Python do you use? If it's less than 3.6 then you will have added complications in both approaches (if the output order is important) — roganjosh
– roganjosh, Commented Aug 8, 2019 at 17:02

Luke · Accepted Answer · 2019-08-08 17:18:26Z

1

new_list = [d[key][0] for key in d]

That's a list comprehension -- it loops throught the keys in d, and then gets the first item of each value, using that key.

answered Aug 8, 2019 at 17:18

Luke

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Luke Over a year ago

you can add conditionals in a comprehension as well -- for instance: new_list = [d[key][0] for key in d if type(key) == int and key % 2 == 0] that'll only grab the even numbered key if it's an integer.

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