Python 从字典更新列表中的值？

Question

I'm trying to convert values in nested list from dict

list1 =  [["a","b"],["c"],["a","d"]]

list2 =  [["a","b"],["c"],[""],["a","d"]]

dict_form = {"a":"app","b":"bold","c":"cafe","d":"dot"}

Expected output

new_list = [["app","bold"],["cafe"],["app","dot"]]

result_1 = [["app","bold"],["cafe"],[""],["app","dot"]]

What I have tried:

result = [[dict_form[i] for i in j] for j in new_list]

Answer 1

Use a 2-level list comprehension

result = [[dict_form[k] for k in sublist] for sublist in list1]
print(result) # [['app', 'bold'], ['cafe'], ['app', 'dot']]

---

Timeit on 20 loops of 200k iterations

# List comprehension : [[dict_form[k] for k in sublist] for sublist in list1]
python -m timeit -r 20 -n 200000 -s "list1 =  [['a','b'],['c'],['a','d']];dict_form = {'a':'app','b':'bold','c':'cafe','d':'dot'}" "[[dict_form[k] for k in sublist] for sublist in list1]"
200000 loops, best of 20: 1.74 usec per loop

# list/map : list(map(lambda x: list(map(dict_form.get, x)), list1))
python -m timeit -r 20 -n 200000 -s "list1 =  [['a','b'],['c'],['a','d']];dict_form = {'a':'app','b':'bold','c':'cafe','d':'dot'}" "list(map(lambda x: list(map(dict_form.get, x)), list1))"
200000 loops, best of 20: 3.2 usec per loop

So more performant on normal list length, and same on huge list (see Derek post)

Answer 2

Here is how you can use a nested list comprehension:

list1 =  [["a","b"],["c"],["a","d"]]
dict_form = {"a":"app","b":"bold","c":"cafe","d":"dot"}
new_list = [[dict_form[k] for k in ls] for ls in list1]
print(new_list)

Output:

[['app', 'bold'], ['cafe'], ['app', 'dot']]

Answer 3

another way would be to just use map:

list(map(lambda x: list(map(dict_form.get, x)), list1))

output:

[['app', 'bold'], ['cafe'], ['app', 'dot']]

slight variation:

[*map(lambda x: [*map(dict_form.get, x)], list1)]

As a follow up to @azro's timings, I generated a much larger input list1 with the same kind of structure..

using the current input from the OP, approx:

nested list comprehension = 0.000000833 s

map method = 0.000001440 s

map method variation = 0.000001180 s

with a MUCH larger but similar list, approx:

nested list comprehension = 0.054890156 s

map method = 0.050899744 s

map method variation = 0.065859318 s

So I guess the most efficient solution (from those provided) depends on the size of your actual list, seems the map method tends to perform the best as the lists get larger