假设我有这个:
[
{"name": "Tom", "age": 10},
{"name": "Mark", "age": 5},
{"name": "Pam", "age": 7}
]
通过搜索“Pam”作为名称,我想检索相关的字典:{name:“Pam”,年龄:7}
如何做到这一点?
假设我有这个:
[
{"name": "Tom", "age": 10},
{"name": "Mark", "age": 5},
{"name": "Pam", "age": 7}
]
通过搜索“Pam”作为名称,我想检索相关的字典:{name:“Pam”,年龄:7}
如何做到这一点?
当前回答
为@FrédéricHamidi添加一点点。
如果你不确定字典列表中是否有键,这样做会有帮助:
next((item for item in dicts if item.get("name") and item["name"] == "Pam"), None)
其他回答
为@FrédéricHamidi添加一点点。
如果你不确定字典列表中是否有键,这样做会有帮助:
next((item for item in dicts if item.get("name") and item["name"] == "Pam"), None)
你必须遍历列表中的所有元素。没有捷径!
除非在其他地方保存了一个包含指向列表项的名称的字典,但这时必须处理从列表中弹出元素的后果。
这里是一个比较,使用遍历列表,使用过滤器+lambda或重构(如果需要或对你的情况有效)你的代码dict of dicts而不是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
输出是这样的:
List of Dictionries took: 0.0099310874939
Using filter took: 0.0121960639954
Dict of Dicts took: 4.05311584473e-06
结论: 显然,字典是最有效的搜索方法在这种情况下,你知道,你只能通过id搜索。 有趣的是,使用过滤器是最慢的解决方案。
我测试了各种方法来遍历字典列表并返回键x具有特定值的字典。
结果:
速度:列表推导>生成器表达式>>常规列表迭代>>>过滤器。 所有的比例都与列表中的字典数量线性(10x列表大小-> 10x时间)。 对于大量(数千个)键,每个字典的键不会显著影响速度。请看我计算的图表:https://imgur.com/a/quQzv(方法名称见下文)。
所有测试均使用Python 3.6.4, W7x64完成。
from random import randint
from timeit import timeit
list_dicts = []
for _ in range(1000): # number of dicts in the list
dict_tmp = {}
for i in range(10): # number of keys for each dict
dict_tmp[f"key{i}"] = randint(0,50)
list_dicts.append( dict_tmp )
def a():
# normal iteration over all elements
for dict_ in list_dicts:
if dict_["key3"] == 20:
pass
def b():
# use 'generator'
for dict_ in (x for x in list_dicts if x["key3"] == 20):
pass
def c():
# use 'list'
for dict_ in [x for x in list_dicts if x["key3"] == 20]:
pass
def d():
# use 'filter'
for dict_ in filter(lambda x: x['key3'] == 20, list_dicts):
pass
结果:
1.7303 # normal list iteration
1.3849 # generator expression
1.3158 # list comprehension
7.7848 # filter
people = [
{'name': "Tom", 'age': 10},
{'name': "Mark", 'age': 5},
{'name': "Pam", 'age': 7}
]
def search(name):
for p in people:
if p['name'] == name:
return p
search("Pam")