Map values in a List based on custom Dictionary - Python

Question

I have a list of numbers:

a = [4,4,4,4,4,4,4,4,4,........................,4,3,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,4]

I want to transform the values according to a custom dictionary, for example:

cust_dict ={4:'four',1:'one',2:'two',3:'three'}

To get the following:

a= [four,four,four,four,four.....,four, three,two,....]

The only code I have done is with a for loop:

for i in range(len(a)):
   a[i] = cust_dict[a[i]]

Is there a more efficient way (in pure python), thus avoiding for-loop ? For a list of 35k items I took around 4ms with this code.

You can use list comprehension, but it won't give you significant boost. — Olvin Roght, Jul 30 '20 at 10:13
@ArvinKushwaha, `a = list(map(cust_dict.get, a))` would be much better. — Olvin Roght, Jul 30 '20 at 10:14
try `a=list(map(cust_dict.get, a))` and see if it's any faster — Itamar Mushkin, Jul 30 '20 at 10:14
@OlvinRoght dict.get is significantly slower than directly indexing. If you are certain that all the data is "safe" then your best bet is indexing. But yeah, if it isn't "safe", then dict.get is better. — Arvin Kushwaha, Jul 30 '20 at 10:16
4ms isn't slow. If you have lists of (over) 35K, you may want to look at other options to store and use your data. Pandas Series type even has a nice `.map()` method, which should work very straightforwardly. — 9769953, Jul 30 '20 at 10:22
Does this answer your question? [List comprehension vs map](https://stackoverflow.com/questions/1247486/list-comprehension-vs-map) — mkrieger1, Jul 30 '20 at 10:26

score 3 · Answer 1 · answered Jul 30 '20 at 10:23

3

Thank for the mapping, I was looking exactly for something like that. In term of speed (on my list (35k entries)):

List Comprehension (@Arvin Kushwaha): a = [cust_dict[i] for i in a] --> 3 ms
Lambda Mapping(@Arvin Kushwaha) : a = list(map(lambda x: cust_dict[x], a)) --> 5.54ms
Dict get Mapping(@Olvin Roght) : a=list(map(cust_dict.get, a) --> 2ms

PS: pandas mapping took 9ms (with transforming back the pd.series to a list) Thank you all!

answered Jul 30 '20 at 10:23

Pdeuxa

651
7
27

Depending on further needs, you may want to keep things as an array or series, instead of using lists again. Lists can be fast, but at some point, an array or series will outdo a Python list. – 9769953 Jul 30 '20 at 10:27
I agree, if I have additionnal processing behind, to let things in series or arrays. – Pdeuxa Jul 30 '20 at 10:29
1

@Pdeuxa, if you're really obsessed with execution time, you can replace you `cust_dict` with `cust_list = [None, 'one', 'two', 'three', 'four']` and use `list(map(cust_list.__getitem__, a))`, it should be slightly faster. – Olvin Roght Jul 30 '20 at 10:36

score 1 · Answer 2 · answered Jul 30 '20 at 10:22

Have a look at this:

a= [4,4,4,4,4,4,4,4,4,4,3,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,4]

cust_dict ={4:'four',1:'one',2:'two',3:'three'}

output = list(map(lambda x: cust_dict[x], a))

print(output)
# ['four', 'four', 'four', 'four', 'four', 'four', 'four', 'four', 'four', 'four', 'three', 'two', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'four']

9769953 · Accepted Answer · 2020-07-30T10:31:06.047

With 35K items, I would use a NumPy array, or in this context, a Pandas Series (this obviously ignores the "pure Python" mention in your question):

>>> import pandas as pd
>>> a = [4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 4, 4, 4, 1, 1, 1, 1, 2, 2, 2, 3]
>>> cust_dict ={4:'four',1:'one',2:'two',3:'three'}
>>> s = pd.Series(a)
>>> s.map(cust_dict).tolist()
['four', 'four', 'four', 'four', 'four', 'four', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'three', 'two', 'two', 'two', 'four', 'four', 'four', 'one', 'one', 'one', 'one', 'two', 'two', 'two', 'three']

But you may not want to convert the series back to a list, depending on further needs and usage.

score 0 · Answer 4 · answered Jul 30 '20 at 10:26

0

a[:] = map(lambda x:cust_dict[x], a)

answered Jul 30 '20 at 10:26

SID007

21
3

Map values in a List based on custom Dictionary - Python

4 Answers4