how to determine the minimal value in a column of a list of tuples in python

Simplest way, you can use min,

>>> lstoflsts = [(1.2, 2.1, 3.1), ... (0.9, 3.4, 7.4), ... (2.3, 1.1, 5.1)] >>> >>> min(lstoflsts, key=lambda x: x[1]) (2.3, 1.1, 5.1) >>> min(lstoflsts, key=lambda x: x[1])[1] 1.1 

As requested by @udo,

min(zip(*lstoflsts)[1]) 

This will change the list so the columns are rows (rotate it) and then get the 2nd (0 based indexing) row (previously column).

Finally, it returns the minimun value.

While sharing a love of generator expressions (which in this case is both one line and beautifully self-documenting once you know Python), is there anything actually wrong with

m =lstoflsts[0][1] for x in lstoflsts: m = min(m, x[1] ) 

It works exactly the same way you'd sort a list according to that criterion. Pass a key function:

min(iterable[, key]):

[...] The optional key argument specifies a one-argument ordering function like that used for list.sort().

For example using operator.itemgetter():

import operator lstoflsts = [(1.2, 2.1, 3.1), (0.9, 3.4, 7.4), (2.3, 1.1, 5.1)] print min(lstoflsts, key=operator.itemgetter(1)) # prints (2.3, 1.1, 5.1) 

You can also use a lambda expression as key function, of course. However, using operator.itemgetter() is generally considered more efficient than the lambda function. It is especially more efficient than anything involving zip().

For reference:

• https://docs.python.org/2/library/functions.html#min
• https://docs.python.org/2/library/operator.html#operator.itemgetter