Can I use lambda expression to count the elements that I'm interested? For example, when I need to count the elements in a list that is more than two, I tried this code which returns 0.
x = [1,2,3]
x.count(lambda x: x > 2)
3 Answers
Note: "more than" is > ... => is not a valid operator.
Try sum(y > 2 for y in x)
Or, as suggested by @Jochen, to guard against non-conventional nth-party classes, use this:
sum(1 for y in x if y > 2)
4 Comments
johnsyweb
Summing booleans. Nice. +1 for demonstrating that
lambda is not necessary.
Jochen Ritzel
or
sum(1 for y in x if y > 2)
John Machin
@TokenMacGuy: or it could be decimal, or rational, or some weird DIY class; whatever y is, surely
y > 2 should not run amok. I don't see the point of float(y). Yes, some badass class could define comparison operators that return neither True nor 1 for truthy results -- in which case, the suggestion of @Jochen [Thanks!] avoids the explicit bool()
steveha
sum(1 for y in x if y > 2) is IMHO the best and most Pythonic answer.You can try any of the following
len([y for y in x if y > 2])
or
len(filter(lambda y: y > 2, x))
or the nicer
sum( y > 2 for y in x )
4 Comments
John Machin
OP wanted "more than two", not "more than or equal to two".
vijayst
I tried the second option. But it says 'filter' has no len.
donkopotamus
@vijayst This answer is from 2011 and thus for Python 2.7.
filter returns a generator in Python 3.x meaning that suggested method is no longer applicable.
trcm
Second option just needs the filter wrapping in list() to work with Python3 :
len(list(filter(lambda y: y > 2, x)))from functools import reduce
x = [1,2,3]
reduce(lambda a,i: a+1 if (i>2) else a, x, 0)
This will not create a new list. a is the accumulator variable, i is the item from the list, and the 0 at the end is the initial value of the accumulator.
lambdas?