获取字典中的最大键

Question

I have a dictionary that looks like this

MyCount= {u'10': 1, u'1': 2, u'3': 2, u'2': 2, u'5': 2, u'4': 2, u'7': 2, u'6': 2, u'9': 2, u'8': 2}

I need highest key which is 10 but i if try max(MyCount.keys()) it gives 9 as highest.
Same for max(MyCount) .

The dictionary is created dynamically.

Answer 1

This is because u'9' > u'10' , since they are strings .

To compare numerically, use int as a key.

max(MyCount, key=int)

(Calling .keys() is usually unnecessary)

Answer 2

You need to compare the actual numerical values. Currently you're comparing the strings lexigraphically.

max(MyCount, key=int)

Answer 3

max(map(int, MyCount))

或者，如果您希望返回值是原始字符串：

max(MyCount, key=int)

Answer 4

Since your keys are strings, they are compared lexicographically and '9' is the max value indeed.

What you are looking for is something like: max(int(k) for k in MyCount)

Answer 5

This is your problem:

>>> u'10' > u'9'
False

Effectively, you're comparing the characters '1' and '9' here. What you want is probably this:

max(long(k) for k in MyCount)

or create the dictionary with numbers as keys (instead of strings).

Answer 6

You use max for string values. You must convert them to int. Try something like:

print(max([int(s) for s in MyCount.keys()]))

Or as Tim suggested:

print(max(int(s) for s in MyCount))

Answer 7

如果有人最终在这里寻找如何按值而不是键排序：

max(MyCount, key=MyCount.get)