Truncate dictionary list values

Question

I'm trying to find the keys for the matching values in a dictionary. But to get any kind of valid match I need to truncate the values in the lists.

I want to truncate down the the tenths spot (e.g. "2.21" to "2.2").

dict1 = {'red':[1.98,2.95,3.83],'blue':[2.21,3.23,4.2333],'orange':[3.14,4.1,5.22]}

dict2 = {'green':[3.11,4.12,5.2],'yellow':[2.2,3.2,4.2],'red':[5,2,6]}

matches = []
for key1 in dict1:
    for key2 in dict2:
        if dict1[key1] == dict2[key2]:
            matches.append((key1, key2))

print(matches)

I'm trying to get "green" and "orange" should be a match, as well as "blue" and "yellow". But I'm not sure if I need to parse through each value list first, make the changes, and then continue. If I could make the changes in the comparison itself that would be ideal.

Answer 1

d = {'green': [3.11, 4.12, 5.2]}

>>> map(int, d['green'])
[3, 4, 5]

You need to map the list items to integers before you compare

for key2 in dict2:
    if map(int, dict1[key1]) == map(int, dict2[key2]):
        matches.append((key1, key2))

I'm assuming you want to round down. If you want to round to the nearest integer use round instead of int

Answer 2

You can use zip on the lists and compare values, but you should set a tolerance tol for which a pair of values from two list will be considered the same:

dict1 = {'red':[1.98,2.95,3.83],'blue':[2.21,3.23,4.2333],'orange':[3.14,4.1,5.22]}
dict2 = {'green':[3.11,4.12,5.2],'yellow':[2.2,3.2,4.2],'red':[5,2,6]}

matches = []
tol = 0.1 # change the tolerance to make comparison more/less strict
for k1 in dict1:
    for k2 in dict2:
        if len(dict1[k1]) != len(dict2[k2]):
            continue
        if all(abs(i-j) < tol for i, j in zip(dict1[k1], dict2[k2])):
            matches.append((k1, k2))

print(matches)
# [('blue', 'yellow'), ('orange', 'green')]

If your list lengths will always be the same, you can remove the part where non matching lengths are skipped.