Python: Find index of a dictionary from a list of dictionaries

Question

I have a list of dictionaries. How can I find the index of specific dictionary? For example:

[{'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}]

This is the list of dictionaries. I want to find the index of this dictionary from the list:

{'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}

Answer 1

The list type in python is mutable, that means the object itself does not guarantee that order is preserved during transformations. Thus using such an approach is to be done with caution.

With that out of the way you could do sth like

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)

Answer 2

Lists in python3 have an index function that you can use to find the index of specific elements -

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))

Keep in mind that it will only return the first match in the case that there are multiple matches (since lists can have duplicate elements)

Answer 3

If you really need an array inside the dictionary as opposed to a list, you'll need to write a custom comparator, since the equal() method (invoked by the == operator) behaves differently for a numpy.array.

And you would receive the following error:

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

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

Instead of comparing the whole arrays and returning a boolean answer, it returns an element-wise comparison (array):

np.array([1,2,3]) == np.array([1,2,3])
array([ True,  True,  True])

Thus for a single boolean answer in case of an array, we need to call explicitly the array_equal() method instead:

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

Then expanding upon oskros answer :

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