Assume I have this:
[
{\"name\": \"Tom\", \"age\": 10},
{\"name\": \"Mark\", \"age\": 5},
{\"name\": \"Pam\", \"age\": 7}
]
and by searchin
Here is a comparison using iterating throuhg list, using filter+lambda or refactoring(if needed or valid to your case) your code to dict of dicts rather than list of dicts
import time
# Build list of dicts
list_of_dicts = list()
for i in range(100000):
list_of_dicts.append({'id': i, 'name': 'Tom'})
# Build dict of dicts
dict_of_dicts = dict()
for i in range(100000):
dict_of_dicts[i] = {'name': 'Tom'}
# Find the one with ID of 99
# 1. iterate through the list
lod_ts = time.time()
for elem in list_of_dicts:
if elem['id'] == 99999:
break
lod_tf = time.time()
lod_td = lod_tf - lod_ts
# 2. Use filter
f_ts = time.time()
x = filter(lambda k: k['id'] == 99999, list_of_dicts)
f_tf = time.time()
f_td = f_tf- f_ts
# 3. find it in dict of dicts
dod_ts = time.time()
x = dict_of_dicts[99999]
dod_tf = time.time()
dod_td = dod_tf - dod_ts
print 'List of Dictionries took: %s' % lod_td
print 'Using filter took: %s' % f_td
print 'Dict of Dicts took: %s' % dod_td
And the output is this:
List of Dictionries took: 0.0099310874939
Using filter took: 0.0121960639954
Dict of Dicts took: 4.05311584473e-06
Conclusion: Clearly having a dictionary of dicts is the most efficient way to be able to search in those cases, where you know say you will be searching by id's only. interestingly using filter is the slowest solution.