Question

我有一个包含对象的列表，即：

[{"id":"1", "name": "name1"}, {"id":"2", "name": "nam2"}, {"id":"1", "name": "name3"}]

我需要找到具有“ id”：“ 1”的列表中对象的数量，在此示例中：2

Answer 1

如果collections.Counter拥有一个key和sorted这样的itertools.groupby参数，这很好，但是不幸的是没有。

我将使用collections.defaultdict作为计数器。作为奖励，我们将获得所有不同ID的计数器：

from collections import defaultdict


li = [{"id":"1", "name": "name1"}, {"id":"2", "name": "nam2"}, {"id":"1", "name": "name3"}]

counter = defaultdict(int)

for d in li:
    counter[d['id']] += 1

print(counter['1'])
# 2
print(counter['2'])
# 1

更新

我似乎忽略了以下事实：您仍然可以使用collections.Counter作为注释中提到的@RoadRunner：

from collections import Counter


li = [{"id":"1", "name": "name1"}, {"id":"2", "name": "nam2"}, {"id":"1", "name": "name3"}]

print(Counter(d['id'] for d in li))
# Counter({'1': 2, '2': 1})

Answer 2

您可以使用列表推导来过滤所需的数据。例如：

# Data:
l=[{"id":"1", "name": "name1"}, {"id":"2", "name": "nam2"}, {"id":"1", "name": "name3"}]
print(len([x for x in l if x['id']=='1'])) # Result: 2
#         ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
#         This is the key part: you filter the list according to a condition
#         (in this case: x['id']=='1').
#         If all you need is the number of entries for which the condition
#         holds, printing the length of the resulting list will be enough.

Answer 3

对此问题有不同的看法：

import operator

a = [{'id': '1', 'name': 'name1'},
 {'id': '2', 'name': 'nam2'},
 {'id': '1', 'name': 'name3'}]

sum(operator.countOf(i, '1') for i in map(operator.itemgetter('id'), a))

有没有办法在python列表中搜索对象？

3 个答案: