Get index in dictionary, not a key value

Question

I have a dictionary like this:

header= {'f1': 13, 'f2': 7, 'f3': 45};

As you see, header['f2'] = 7 has the minimum value and this item is the second in header (its index is 1).

What I do?

I have try this code to get the index of minimum item in header (here 1) but it returns the key value:

index = min(header)

Output:

f2

What I want?

I want to get the index of the minimum item in the dictionary, How can i do that?

Answer 1

Dictionaries do not have indices

Even in Python 3.6+ (officially 3.7+), where dictionaries are insertion ordered , you cannot extract a key or value directly by position. The same is true for collections.OrderedDict . See also: Accessing dictionary items by position in Python 3.6+ efficiently .

min + enumerate

Assuming Python 3.6+, you can extract the position of a key / value based on insertion ordering via iteration. In this case, you can use min with a custom lambda function:

header = {'f1': 13, 'f2': 7, 'f3': 45}

min_idx, (min_key, min_val) = min(enumerate(header.items()), key=lambda x: x[1][1])

print((min_idx, min_key, min_val))

(1, 'f2', 7)

Answer 2

Using header.values() returns a list of just the values of your dictionary:

>>> header= {'f1': 13, 'f2': 7, 'f3': 45}
>>> header.values()
[13, 7, 45]
>>> print(min(header.values()))
7

EDIT: Sorry, you wanted the corresponding key. Here's one way to do it, without having to include any other special libraries:

print(header.keys()[header.values().index(min(header.values()))])

Although dictionaries are not ordered in Python, lists are, and the lists you get from .keys() and .values() line up with each other.

Answer 3

list(header.values()).index(min(header.values()))

这将返回字典中最小值的索引。