How can I simplify 2 for loops to get a dict object value?

Question

Would there be a way to make the two loops simpler, faster, and cleaner, such as List Comprehension,o(n),etc.

def main:
  a={"res":
       [
        {"group":1,"ins":[{"name":"aa","id":123]},
        {"group":2,"ins":[{"name":"cc","id":456]}
     ]}
   result=[]
  for key in a["res"]:
     for i in key["ins"]
        result.append(i)
  return result

Are you primarily trying to make the code more Pythonic or make it run faster? Also, typo: `"re"` or `"res"`. — jarmod, Jan 28 '22 at 16:57

score 4 · Accepted Answer · answered Jan 28 '22 at 16:58

4

Try this:

result = [i for key in a['res'] for i in key['ins']]

answered Jan 28 '22 at 16:58

Riccardo Bucco

13,980
4
22
50

Multivac · Answer 2 · 2022-01-28T17:12:34.193

0

Try:

import numpy as np

# 3.21 µs ± 23.9 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)    
np.ravel([z["ins"] for z in a["res"]]).tolist()

Outputs:

[{'name': 'aa', 'id': 123}, {'name': 'cc', 'id': 456}]

Edit:

You could also try a faster option:

# 372 ns ± 11.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
sum([z["ins"] for z in a["res"]],[])

edited Jan 28 '22 at 17:12

answered Jan 28 '22 at 17:05

Multivac

735
6
10

Would be interesting to add the load/import time of numpy into this timing. – jarmod Jan 28 '22 at 17:23
Oh good point, the first option does not consider the import for the timing so both are comparable – Multivac Jan 28 '22 at 17:25

How can I simplify 2 for loops to get a dict object value?

2 Answers2