confused about max() built in function

Question

hello I am kind of confused about the max() function, I have the following code:

a = '9:30'
b = '10:44'
c = '8:22'
x = max(a, b, c)
print (x)

so my question is: why does this return 9:30? and if I were to delete the a from inside max it would return 8:22

As @chepner mentions above, you may want to look into using `time` rather than strings. see https://docs.python.org/2/library/time.html — bentank, May 02 '15 at 18:53

score 5 · Accepted Answer · answered May 02 '15 at 18:52

5

String a compares as the biggest because it starts with 9 and the other strings start with 1 and 8. Python therefore returns '9:30' as the maximum value out of those three strings.

answered May 02 '15 at 18:52

Simeon Visser

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score 2 · Answer 2 · edited May 23 '17 at 12:06

2

As you are doing string comparison the int or ASCII value of '9' is greater than '8' and '1'. You can see the int value of individual char using ord built-in function.

>>> ord('9')
57
>>> ord('1')
49
>>> ord('8')
56

And as Python does comparison in lexicographical manner(from left to right each char). It finds value of '9' greater. And if you delete '9' if finds '8''s value greater than '1'.

One way to achieve could be :-

a = '9:30'
b = '10:44'
c = '8:22'
d = '10:55'

print max([a, b, c, d], key=lambda x: map(int, x.split(':')))

edited May 23 '17 at 12:06

Community

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answered May 02 '15 at 18:54

Tanveer Alam

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Better remove the `[0]` or it will compare `10:30` and `10:45` as equal. – tobias_k May 02 '15 at 19:14
Also, `lambda i: int(i)` is the same as just `int`. – tobias_k May 02 '15 at 19:40
Yeah but it will return a list of int value. [10,45] and [10, 55] which will return [10, 55] as greater value. – Tanveer Alam May 02 '15 at 19:44

score 1 · Answer 3 · answered May 02 '15 at 18:56

1

Its because of that max function will find the max value lexicographicaly , you can use a proper key for max function to convert the hour and min to int.

>>> l=[a,b,c]
>>> max(l,key=lambda d :map(int, d.split(":")))
'10:44'

answered May 02 '15 at 18:56

Mazdak

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score 0 · Answer 4 · answered May 02 '15 at 18:53

0

Since it's arguments are strings, it is returning the string that is latest in alphabetic order.

answered May 02 '15 at 18:53

Craig Burgler

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score 0 · Answer 5 · answered May 02 '15 at 18:54

it is comparing strings, and, as such, it starts by comparing the first character, and then, if that character is the same, moves on to the next and then the next etc etc.

So '9:30' is returned because 9 > 8 > 1. If you want "10:44" to return, then you should use the datetime module, unless you want to write custom code for that

edit: and sniped by people who type faster!

score 0 · Answer 6 · answered May 02 '15 at 18:56

You are comparing strings, not numbers or times. As such, it compares the strings by their first character (and if those are equal, by the second, and so on), and the first character of "9:30" is the highest.

If you want to compare the times, you can split the string at the :symbol, map the parts to integers and compare those, using an according key function.

x = max([a, b, c], key=lambda s: map(int, s.split(":")))

score 0 · Answer 7 · answered May 02 '15 at 19:01

Here's another way to do it so you can just use the regular max function. First, pad the strings so that "9" becomes "09" and "10" stays "10". You can use str.zfill for this:

a = '9:30'
b = '10:44'
c = '8:22'
a, b, c = map(lambda s: s.zfill(5), (a, b, c))

Now, they will be what you want lexicographically!

print(max([a, b, c]))
10:44

You only have to pad them once, but you will be stuck with:

print(min([a, b, c]))
08:22

confused about max() built in function

7 Answers7