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AlexMoshi
AlexMoshi

Reputation: 339

Python - How to use map function and lambda function with if condition

I want to compare the wordlist and the checklist by using map function and lambda function if the words in wordlist and checklist. this words will be deleted in wordlist. Anyone can help me? thanks a lot

wordlist = ['hello','world','Tom']
checklist = ['hello','world']

print('before')
print(wordlist)

list(map(lambda x: wordlist.remove(x) if x in checklist else None ,wordlist))
print('after')
print(wordlist)

Current Result: before ['hello', 'world', 'Tom'] after ['world', 'Tom']

Expected result: before ['hello', 'world', 'Tom'] after ['Tom']

Upvotes: 1

Views: 1408

Answers (3)

Masklinn
Masklinn

Reputation: 42312

map always produces one output item for each input item, it can not remove elements. Furthermore, map should not be used to mutate objects, that's not its job, and because it's lazy the results can be unexpected.

filter is designed to create an output with less elements than the input, although it's mostly useful if you already have a ready-made predicate (filtering) function.

Since you do not, you can and should use comprehensions which provide a relatively terse way to perform iteration, filtering, mapping and collection in a single construct:

wordlist = ['hello','world','Tom']
checklist = ['hello','world']

print('before')
print(wordlist)

wordlist = [word for word in wordlist if word not in checklist]
print('after')
print(wordlist)

ps: if you want to modify things in-place, use a regular loop

Upvotes: 2

Tom Karzes
Tom Karzes

Reputation: 24052

The problem is you're removing elements from wordlist while iterating over it, which interferes with the iteration. But map is the wrong thing to use for this.

The closest solution to what you're doing now uses filter:

x = list(filter(lambda x: x not in checklist, wordlist))

This returns a filter object which is then converted to a list.

It's probably better to just use a list comprehension:

y = [x for x in wordlist if x not in checklist]

These both produce the same result.

Note that if checklist is large, it would be more efficient to create a set first, then check for membership in the set, e.g. checkset = set(checklist), then x not in checkset. Functionally it's the same, but the set is more efficient if checklist is large.

Upvotes: 1

Varun Narayanan
Varun Narayanan

Reputation: 343

You can try this instead, which doesn't use lambda but accomplishes your goal. Let us know if you absolutely must use lambda. The issue with your lambda expression is that your modifying the list that you are providing to map in the lambda function.

wordlist_2 [word for word in wordlist if word not in checklist and word]

The last and word is to not add None to your list.

Upvotes: 1

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