Python：从字典列表中查找字典索引

我有一份字典清单。如何找到特定词典的索引？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}]

这是字典目录。我想从列表中查找此词典的索引：

{'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

列表

类型是可变的，这意味着对象本身不能保证在转换过程中保持顺序。因此，应谨慎使用这种方法

有了这些，你就可以做类似的事情了

input_list=[{'pop'：[1,0,1,1,1,1,0,0,0,0,0,1,1,1,1,1,1,0,0,1,0，
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],
“重量”：260.1，
“值”：2313}，
{'pop'：[1,1,0,1,1,1,0,1,0,1,0,0,0,0,0,1,0,0,0,0,0,0,1,0，
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],
“重量”：235.6000000000002，
“值”：2774}，
{'pop'：[0,1,1,0,1,1,0,0,0,0,0,0,1,1,1,0,0,0,1,1,1,0,0,1,1,0,0，
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],
“权重”：192.0，
“值”：2254}，
{'pop'：[0,0,1,1,1,1,1,0,0,0,0,0,1,1,1,1,0,0,1,1,1,1,0,1，
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],
“重量”：264.29999995，
“值”：2813}]
elem={'pop'：[0,1,1,1,0,1,1,0,0,0,0,0,1,1,1,0,0,0,1,1,1,0,0,0，
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],
“权重”：192.0，
“值”：2254}
out=next（（i代表i，x在枚举（输入列表）中，如果x==elem），“未找到元素”）
打印（输出）

python3中的列表有一个

索引

函数，您可以使用该函数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，它只会在存在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，因为该方法（由==运算符调用）numpy.array的行为不同

您将收到以下错误：

ValueError: The truth value of an array with more than one element is ambiguous. 

Use a.any() or a.all()

它不会比较整个数组并返回布尔答案，而是返回元素级比较（数组）：

因此，对于数组中的单个布尔答案，我们需要显式调用该方法：

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

然后扩展到oskros：

---

我想这也许可以回答你的问题：你有没有尝试过你的list.index（IteemYouSearch）？

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

out = next((i for i, x in enumerate(dict_list) if elements_equal(x, e)), "Element not found")
print(out)