Python String filter

Question

I am trying to solve a class assignment as listed below

Use lambda expressions and the filter() function to filter out words from a list that don't start with the letter 's'. For example:

seq = ['soup','dog','salad','cat','great']

should be filtered down to:

['soup','salad']

I wrote the following two sets of code...

seq = ['soup', 'dog','salad', 'cat', 'great']
result = filter (lambda test: test[0] == "s",seq)
print (result)

seq = ['soup', 'dog','salad', 'cat', 'great']
def checkf (input):
    if input[0]=='s':
            a = "TRUE"
    return(a)
test = seq[0]
result = filter (checkf(test),seq)
print (result)

Both codes give a strange answer filter object at 0x112aa06d8 as if it is returning a memory address in hexadecimal... What's wrong?

I have edited your question to more appropriately reflect a more readable question see stackoverflow.com/help/how-to-ask for further help in regards to asking — Tyler Cowan
– Tyler Cowan, Commented Jan 25, 2018 at 23:55
Possible duplicate of Why does foo = filter(...) return a <filter object>, not a list? — cowbert
– cowbert, Commented Jan 25, 2018 at 23:58
In Python 3, filter returns a special filter object rather than a list. You can convert it to a list using the list function. — khelwood
– khelwood, Commented Jan 25, 2018 at 23:58
what is wrong with the first code. Works for me with python 2. — Monish K Nair
– Monish K Nair, Commented Jan 26, 2018 at 0:00
Possible duplicate of How to use filter, map, and reduce in Python 3 — melwil
– melwil, Commented Jan 26, 2018 at 0:00

melwil · Accepted Answer · 2018-01-25 23:59:27Z

3

In python 3, filter objects are iterators. To print the contents, you have to materialize it into a sequence first, so try:

print( list(result))

See also: How to use filter, map, and reduce in Python 3

edited Jan 25, 2018 at 23:59

melwil

2,5531 gold badge21 silver badges34 bronze badges

answered Jan 25, 2018 at 23:57

cowbert

3,4833 gold badges27 silver badges37 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Leda Grasiele · Accepted Answer · 2019-01-10 00:52:57Z

0

To use lambda and filter to resolve this problem, you need to specify that the result of your code will be a list.

So the following will work:

seq = ['soup', 'dog','salad', 'cat', 'great']
list(filter (lambda test: test[0] == "s",seq))

or (assigning to an object):

seq = ['soup', 'dog','salad', 'cat', 'great']
result = list(filter (lambda test: test[0] == "s",seq))
print(result)

The result of the two codes will be:

['soup', 'salad']

edited Jan 10, 2019 at 0:52

answered Jan 10, 2019 at 0:41

Leda Grasiele

4331 gold badge7 silver badges15 bronze badges

Comments

cnnr · Accepted Answer · 2019-09-09 18:01:58Z

0

This is the below code where your seq can be solved according to the lambda function.

seq = ['soup', 'dog','salad', 'cat', 'great']
result = list(filter (lambda test: test[0] == "s",seq))
print(result)

output: ['soup', 'salad']

using it in a function

seq = ['soup', 'dog','salad', 'cat', 'great']
def check(input):
    result = list(filter (lambda test: test[0] == "s",seq))
    return result

check(seq)

output: ['soup', 'salad']

edited Sep 9, 2019 at 18:01

cnnr

1,3074 gold badges18 silver badges23 bronze badges

answered Sep 9, 2019 at 16:14

Debi

166 bronze badges

Collectives™ on Stack Overflow

Python String filter

3 Answers 3

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related