Python: Lambda function

Question

I'm relatively new to Lambda functions and esp this one got me quite confused. I have a set of words that have an average length 3.4 and list of words = ['hello', 'my', 'name', 'is', 'lisa']

I want to compare the length of a word to average length and print out only the words higher than average length

average = 3.4
words = ['hello', 'my', 'name', 'is', 'lisa']

print(filter(lambda x: len(words) > avg, words))

so in this case i want

['hello', 'name', 'lisa']

but instead I'm getting:

<filter object at 0x102d2b6d8>

Padraic Cunningham · Accepted Answer · 2014-11-06 22:34:26Z

3

list(filter(lambda x: len(x) > avg, words)))

filter is an iterator in python 3. You need to iterate over or call list on the filter object.

In [17]: print(filter(lambda x: len(x) > avg, words))
<filter object at 0x7f3177246710>

In [18]: print(list(filter(lambda x: len(x) > avg, words)))
['hello', 'name', 'lisa']

answered Nov 6, 2014 at 22:34

Padraic Cunningham

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Daniel · Accepted Answer · 2014-11-06 22:33:07Z

3

lambda x: some code

Means that x is the variable your function gets so basically you need to replace words with x and change avg to average because there isn't a variable called avg in your namespace

print filter(lambda x: len(x) > average, words)

answered Nov 6, 2014 at 22:33

Daniel

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Stuart · Accepted Answer · 2014-11-06 22:37:54Z

2

In case you're wondering why there are contradictory answers being given here, note that in Python 2 you can do:

print filter(lambda x: len(x) > avg, words)

But as others have said, in Python 3 filter is an iterator and you have to iterate through its contents somehow to print them.

answered Nov 6, 2014 at 22:37

Stuart

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kindall · Accepted Answer · 2014-11-06 22:43:00Z

2

Others have covered your problems with the lambda. Now this:

<filter object at 0x102d2b6d8>

This is your result. It is not an error; it is an iterator. You have to use it as a sequence somehow. For example:

print(*filter(lambda x: len(x) > average, words))

Or if you want it as a list:

print(list(filter(lambda x: len(x) > average, words)))

edited Nov 6, 2014 at 22:43

answered Nov 6, 2014 at 22:36

kindall

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2 Comments

Padraic Cunningham Over a year ago

but this will not return a list

kindall Over a year ago

print won't return anything anyway. But if you want it as a list, just pass it to list(). I've added that example.

heltonbiker · Accepted Answer · 2014-11-06 22:35:52Z

1

You need to have same variable in both sides of : operator.

lambda x:len(x) > avg

Here, x is the argument. This is equivalent to:

def aboveAverage(x):
    return len(x) > avg

only without aboveAverage function name.

Example use:

print map(lambda x:len(x) > avg, words)

edited Nov 6, 2014 at 22:35

answered Nov 6, 2014 at 22:30

heltonbiker

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