Can I return a list of ALL min tuples, using Python's Min function?

Question

Say I have a list of tuples, like the following:

listo = [('a','1'),('b','0'),('c','2'),('d','0')]

If I want the lowest tuple, based on the second index of each tuple, I can customize the min function with a lambda function, like this:

min(listo, key=lambda x: x[1])

As it stands, this code would return:

In [31]: min(listo, key=lambda x: x[1]) Out[31]: ('b', '0')

But this only gives me one tuple, and only the first one it encounters at that. What if I wanted ALL the min tuples? So it returns something like:

In [31]: min(listo, key=lambda x: x[1]) Out[31]: [('b', '0'),('d','0')]

Any advice on how to accomplish this?

Disconnect3d · Accepted Answer · 2013-06-27 20:35:58Z

You can do it by first searching one minimum value and then look for another ones based by the minimum value found.

listo = [('a','1'),('b','0'),('c','2'),('d','0')] minValue = min(listo, key=lambda x: x[1])[1] minValueList = [x for x in listo if x[1] == minValue]

Another approach could be making your own minimum function that uses some kind of list of minimum values but probably it would be much slower for bigger problems.

Nice, I was kind of doing the second approach already, but was looking for a more "Pythonic" way. I like this, thank you.

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