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[英]Created multi index nested json or dictionary from list of dictionaries in Python
[英]Python: Find index of a dictionary from a list of dictionaries
我有一个字典列表。 如何找到特定字典的索引? 例如:
[{'pop': array([1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0,
1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1,
1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0,
1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1,
0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1]),
'weight': 260.1,
'value': 2313},
{'pop': array([1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0,
0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1,
0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1,
0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0,
0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1]),
'weight': 235.60000000000002,
'value': 2774},
{'pop': array([0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0,
0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1,
1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0,
0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1,
1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1]),
'weight': 192.0,
'value': 2254},
{'pop': array([0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1,
1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1,
0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0,
1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1,
0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1]),
'weight': 264.29999999999995,
'value': 2813}]
这是字典列表。 我想从列表中找到这本词典的索引:
{'pop': array([0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0,
0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1,
1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0,
0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1,
1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1]),
'weight': 192.0,
'value': 2254}
python 中的list
类型是可变的,这意味着对象本身不保证在转换过程中保持顺序。 因此,使用这种方法要谨慎。
这样你就可以做某事
input_list = [{'pop': [1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0,
1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1,
1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0,
1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1,
0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1],
'weight': 260.1,
'value': 2313},
{'pop': [1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0,
0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1,
0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1,
0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0,
0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1],
'weight': 235.60000000000002,
'value': 2774},
{'pop': [0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0,
0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1,
1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0,
0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1,
1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1],
'weight': 192.0,
'value': 2254},
{'pop': [0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1,
1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1,
0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0,
1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1,
0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1],
'weight': 264.29999999999995,
'value': 2813}]
elem = {'pop': [0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0,
0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1,
1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0,
0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1,
1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1],
'weight': 192.0,
'value': 2254}
out = next((i for i, x in enumerate(input_list) if x == elem), "Element not found")
print(out)
python3 中的列表有一个index
函数,你可以用它来查找特定元素的索引 -
my_list = [{'pop': [1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0,
1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1,
1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0,
1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1,
0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1],
'weight': 260.1,
'value': 2313},
{'pop': [1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0,
0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1,
0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1,
0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0,
0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1],
'weight': 235.60000000000002,
'value': 2774},
{'pop': [0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0,
0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1,
1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0,
0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1,
1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1],
'weight': 192.0,
'value': 2254},
{'pop':[0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1,
1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1,
0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0,
1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1,
0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1],
'weight': 264.29999999999995,
'value': 2813}]
my_dict = {'pop': [0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0,
0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1,
1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0,
0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1,
1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1],
'weight': 192.0,
'value': 2254}
print(my_list.index(my_dict))
请记住,它只会在有多个匹配项的情况下返回第一个匹配项(因为列表可以有重复的元素)
如果您确实需要字典中的数组而不是列表,则需要编写自定义比较器,因为equal()方法(由 == 运算符调用)对于 numpy.array 的行为不同。
你会收到以下错误:
ValueError: The truth value of an array with more than one element is ambiguous.
Use a.any() or a.all()
它不是比较整个数组并返回一个布尔值,而是返回一个逐元素比较(数组):
np.array([1,2,3]) == np.array([1,2,3])
array([ True, True, True])
因此,对于数组的单个布尔答案,我们需要显式调用array_equal()方法:
import numpy as np
def elements_equal(x, y):
if x is None and y is None: return True
if x is not None and y is not None:
if x.keys() != y.keys(): return False
for key in x.keys():
if type(x[key]) is np.ndarray:
if not np.array_equal(x[key], y[key]): return False
elif x[key] != y[key]: return False
return True
return False
然后扩展 oskros 答案:
out = next((i for i, x in enumerate(dict_list) if elements_equal(x, e)), "Element not found")
print(out)
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