CODEWITHSUNDEEP
pythonlistdictionarylist-comprehension
macintosh81
macintosh81

Reputation: 189

List comprehensions of dictionary values

I am new to Python and I am trying to learn manipulating a dictionary. I have a dict that has the following structure:

dict = {'city1':([1990, 1991, 1992, 1993],[1.5,1.6,1.7,1.8]),
        'city2':([1993, 1995, 1997, 1999],[2.5,3.6,4.7,5.8])

I would like convert the key of in the following way

 'city': ([1990, 1.5],[1991, 1.6],[1992,1.7],[1993,1.8])

I tried using a for loop to loop through the values and create a new value for each key. However, this seems really slow and clumsy. Is there any Pathonic way of achieving this goal?

Thanks!

Upvotes: 4

Views: 1620

Answers (5)

ettanany
ettanany

Reputation: 19806

One approach would be using this concise solution:

res = {k: zip(v[0], v[1]) for k, v in d.items()}

Output:

>>> {k: zip(v[0], v[1]) for k, v in d.items()}
{'city2': [(1993, 2.5), (1995, 3.6), (1997, 4.7), (1999, 5.8)], 'city1': [(1990, 1.5), (1991, 1.6), (1992, 1.7), (1993, 1.8)]}

Upvotes: 1

cs95
cs95

Reputation: 402263

This gives you exactly the output you want:

d = { k : tuple(map(list, zip(*d[k]))) for k in d }

Output:

{'city2': ([1993, 2.5], [1995, 3.6], [1997, 4.7], [1999, 5.8]), 'city1': ([1990, 1.5], [1991, 1.6], [1992, 1.7], [1993, 1.8])}

Also, do consider a different name than dict as this is the name of the built-in dict class.

zip(*d[k]) is a simplified version of zip(d[k][0], d[k][1]) but they both do the same thing and generate pairwise tuples.

map(list, ...) generates a map object that converts each element in the zip to a list (in python2 tuples are automatically converted to lists)

tuple(...) converts the map list/generator to a tuple of lists, which is exactly what you want.

Upvotes: 2

Ajax1234
Ajax1234

Reputation: 71451

You can try this:

d = {'city1':([1990, 1991, 1992, 1993],[1.5,1.6,1.7,1.8]),
    'city2':([1993, 1995, 1997, 1999],[2.5,3.6,4.7,5.8])}   

new_dict = {a:tuple(map(list, zip(b[0], b[1]))) for a, b in d.items()}

Output:

{'city2': ([1993, 2.5], [1995, 3.6], [1997, 4.7], [1999, 5.8]), 'city1': ([1990, 1.5], [1991, 1.6], [1992, 1.7], [1993, 1.8])}

Upvotes: 3

Alter
Alter

Reputation: 3464

I'm just going to throw my answer in, because I haven't seen anyone take advantage of for key, (year, value) yet. Which I find quite readable and direct

d = {'city1':([1990, 1991, 1992, 1993],[1.5,1.6,1.7,1.8]),
    'city2':([1993, 1995, 1997, 1999],[2.5,3.6,4.7,5.8])} 

a = {k : [list(group) for group in zip(year, val)] for k, (year, val) in d.items()}

enter image description here

Upvotes: 1

Mr_U4913
Mr_U4913

Reputation: 1344

You can try the following code

a = [[a, b] for a, b in zip(dict['city1'][0], dict['city1'][1])]

Output

[[1990, 1.5], [1991, 1.6], [1992, 1.7], [1993, 1.8]]

Upvotes: 3

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