Python-获取元组字典中特定元组索引最小值的键

Question

I have a dict of tuples such as:

d = {'a': (3, 5), 'b': (5, 8), 'c': (9, 3)}  

I want to return the key of the minimum of the tuple values based on the tuple index. For example, if using tuple index = 0, then 'a' would be returned. if index = 1, then 'c' would be returned. I have tried using min(), for example

min(d, key=d.get)

but am not sure how to manipulate it to select the tuple index to use. Although there are similar questions, I have not found an answer to this. Apologies in advance if this is a duped question, and please link to the answer. Thanks

Answer 1

You can write a lambda function to get the elements from the value by their index:

min(d, key=lambda k: d[k][0])
# 'a'
min(d, key=lambda k: d[k][1])
# 'c'

Answer 2

Since multiple keys could have the same value, you might want to return a list of matching keys, not just a single key.

def min_keys(d, index):

    # Initialize lists
    values = []
    matches = []

    # Append tuple items to list based on index
    for t in list(d.values()):
        values.append(t[index])

    # If the item matches the min, append the key to list
    for key in d:
        if d[key][index] == min(values):
            matches.append(key)

    # Return a list of all keys with min value at index
    return matches

Answer 3

Dictionaries are unsorted and have no index.

If you want the return the key alphabetically first you could use the ascii order:

print(chr(min([ord(key) for key in d.keys()])))

Answer 4

Here's a portable method you can use for dicts with a structure like yours, and feel free to choose the index of interest in the tuple:

def extract_min_key_by_index(cache, index):
    min_val = float('inf')
    min_key = 0

    for k, v in d.iteritems():
        if v[index] < min_val:
            min_key, min_val = k, v[index]

    return min_key


d = {'a': (3, 5), 'b': (5, 8), 'c': (9, 3)}
INDEX = 0

print extract_min_key_by_index(d, INDEX)