How to use min() only with first column of 2-dimensional list

Question

I have 2-dimensional list in which each entry has a timestamp (int) and a second value that can be an int, string or object.

I'm using the min() function to get the entry with the lowest timestamp:

a = [[2, "start C"],
     [6, 2484],
     [15, "finish C"]]

print(min(a))

Output:

[2, 'start C']

When there are identical lowest timestamps min() will compare the second column of those items.

This works only as long as the second column has comparable values:

a = [[2, "start C"],
     [2, "start B"],
     [15, "finish C"]]

print(min(a))

Output:

[2, 'start B']

If the second column contains different data types it throws an error:

a = [[2, "start C"],
     [2, 2484],
     [15, "finish C"]]

print(min(a))

Output:

TypeError: '<' not supported between instances of 'int' and 'str'

How can I let only the first column be checked for the minimum? When there are identical timestamps it should simply return the first item encountered.

score 4 · Answer 1 · answered Sep 30 '19 at 14:13

4

You can pass a comparison key

min(my_list, key=lambda x: x[0])

By default, python uses the first item for comparison, and in case of tie, the second one as tie-breaker (and so on), that's why you get the error.

answered Sep 30 '19 at 14:13

blue_note

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yatu · Accepted Answer · 2019-09-30T14:24:29.460

3

min accepts a key argument, which can be used to obtain the expected behaviour:

from operator import itemgetter

a = [[2, "start C"],
     [6, 2484],
     [15, "finish C"]]

min(a, key=itemgetter(0))
# [2, 'start C']

Which is equivalent to slicing the inner lists with a custom lambda function as in @blue_note's solution, but typically somewhat faster (though its mainly a habit from being more convenient when having to fetch multiple items as explained here):

min(my_list, key=lambda x: x[0])

By using itemgetter, only the first element in each inner list will be fetched for ordering purposes, and then the minimum will be computed among this subset

edited Sep 30 '19 at 14:24

answered Sep 30 '19 at 14:14

yatu

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You should probably also mention that this is equivalent to the `lambda` counterpart, but typically significantly faster. – norok2 Sep 30 '19 at 14:22
The timing difference is a bit higher than what the wording of the most accepted answer from [there](https://stackoverflow.com/a/11287216/9698684) seemed to indicate. See my [benchmarks](https://stackoverflow.com/a/58171096/5218354) for some numbers for both `sorted()` and `min()`. My understanding is that with the `lambda` version there is an extra function call implicit to the `[]` operator. – norok2 Sep 30 '19 at 15:53

How to use min() only with first column of 2-dimensional list

2 Answers2