How to find the max number(s) in a list with tied

问题:

So say I have a list like:

my_list = [12, 13, 51, 21, 22, 58, 45.1, 34.2, 56, 6, 58, 58] 

So the max number in this is obviously 58, but I don't just want to return one 58, I want a list of all the indexes that have that max number.

Basically for this I want the result [5, 10, 11]

I know that if I want the max number I can do my_list.index(max(my_list)) but that will simply give me the first index.

Any tips? Also, I want to stick to simple methods such as sort, max, len, etc...

回答1:

You can determine the maxval with max:

maxval = max(my_list)

Then get the indices using enumerate and a list comprehension:

indices = [index for index, val in enumerate(my_list) if val == maxval]

For your example, I get

maxval == 58
indices = [5, 10, 11]

---

As per Keyser's suggestion, you could save iterating over the list twice (once to determine maxval, once to find matching indexes) by doing:

maxval = None
for index, val in enumerate(my_list):
    if maxval is None or val > maxval:
        indices = [index]
        maxval = val
    elif val == maxval:
        indices.append(index)

回答2:

Similar to enumerate and list comprehension, you can also use filter:

maxval = max(my_list)
indices = list(filter(lambda x: my_list[x]==maxval, list(range(len(my_list)))))

回答3:

I tried to do the most Pythonic way as possible

my_list = [12, 13, 51, 21, 22, 58, 45.1, 34.2, 56, 6, 58, 58] 
max_indices = [i for i in range(len(my_list)) if my_list[i] == max(my_list)]

---

Edit: Ashwini is right. max() needs to be outside the list, so...

max_value = max(my_list)
max_indices = [i for i in range(len(my_list)) if my_list[i] == max_value]

回答4:

Enumerate will help you to get the value and location

maxx = max(my_list)
for i,j in enumerate(my_list):
    if j == maxx:
        print i,j

output

5 58

10 58

11 58