CODEWITHSUNDEEP
pythonlistpython-2.7dictionary
Nobi
Nobi

Reputation: 1109

Printing the values of a dictionary when its key is equal to the value of a list element

I have this code below and I want to print the combo[values] for any combo[keys] in combo that are equal to numb[i]

numb = [5, 7, 49, 11, 13]    
combo  = {45 : (-1002,-1023), 49 : (-9999,-2347), 20 : (-1979, -1576), 13 : (-6000,-3450), 110     : (-2139, -8800), 7 : (-6754,-9087) }

How do I do it, please?

Upvotes: 0

Views: 997

Answers (2)

Martijn Pieters
Martijn Pieters

Reputation: 1124070

You mean loop through numb and print any key if present?

Two options; with a loop:

for key in numb:
    if key in combo:
        print combo[key]

which can be expressed as a list comprehension too, to produce a list:

[combo[key] for key in numb if key in combo]

Or with dictionary views:

for key in combo.viewkeys() & numb:
    print combo[key]

again as a list comprehension too:

[combo[key] for key in combo.viewkeys() & numb]

Demo:

>>> numb = [5, 7, 49, 11, 13]    
>>> combo  = {45 : (-1002,-1023), 49 : (-9999,-2347), 20 : (-1979, -1576), 13 : (-6000,-3450), 110     : (-2139, -8800), 7 : (-6754,-9087) }
>>> [combo[key] for key in numb if key in combo]
[(-6754, -9087), (-9999, -2347), (-6000, -3450)]
>>> [combo[key] for key in combo.viewkeys() & numb]
[(-9999, -2347), (-6000, -3450), (-6754, -9087)]

What route you take depends on the size of combo and numb, and on whether numb could be a set as well. If numb could be a set, the dict.viewkeys() could optimize the intersection operation by using the smaller of the two and will most probably be the faster option, especially for larger datasets.

Upvotes: 5

anon582847382
anon582847382

Reputation: 20391

This can be done with a simple list comprehension:

>>> [v for k, v in combo.items() if k in numb]
[(-6754, -9087), (-6000, -3450), (-9999, -2347)]

I think that is what you meant?

Upvotes: 1

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