Python 用 lambda 删除一个字符

Question

I want to remove “X” form this string. What is wrong with my code?

garbled = "IXXX aXXmX aXXXnXoXXXXXtXhXeXXXXrX sXXXXeXcXXXrXeXt mXXeXsXXXsXaXXXXXXgXeX!XX"
message=filter(lambda x: x=="X" x=='' ,garbled)
print message

I know that this one works:

garbled = "IXXX aXXmX aXXXnXoXXXXXtXhXeXXXXrX sXXXXeXcXXXrXeXt mXXeXsXXXsXaXXXXXXgXeX!XX"
message=filter(lambda x: x!="X" ,garbled)
print message

Answer 1

You have the mistake in lambda-condition. You even do not need string.replace() . Lambda-condition must have this syntax:

Use the syntax lambda input: true_return if condition else false_return to return true_return if condition is True and false_return otherwise. condition can be an expression involving input. ( There are more examples )

garbled = "IXXX aXXmX aXXXnXoXXXXXtXhXeXXXXrX sXXXXeXcXXXrXeXt mXXeXsXXXsXaXXXXXXgXeX!XX"
f = lambda x: "" if x in "X" else x
message = filter(f, garbled)
"".join(message)

Answer 2

filter() is a built-in function of Python. The documentation says:

filter(function, iterable)

So you need two things to call the filter function: another function and basically a list of things.

For iterable , a string is fine, because a string is a list of characters. That means, Python can perform a loop on it. That's what's important for an iterable.

For function , let's read on:

Construct an iterator from those elements of iterable for which function returns true. iterable may be either a sequence, a container which supports iteration, or an iterator. If function is None, the identity function is assumed, that is, all elements of iterable that are false are removed.

So it says you need a function that returns True for those items that shall be iterated. For all items you don't want, return False .

Such a function could be:

def remove_all_X(s):
    if s == "X":
        return False  # because you want to remove all "X"
    return True  # all other characters

This can be inversed to

def remove_all_X(s):
    if s != "X":
        return True  # non-"X"
    return False  # "X"

And then shortened to

def remove_all_X(s):
    return s != "X"

You can then use it like

garbled = "IXXX aXXmX aXXXnXoXXXXXtXhXeXXXXrX sXXXXeXcXXXrXeXt mXXeXsXXXsXaXXXXXXgXeX!XX"
message=filter(remove_all_X ,garbled)
print message

Note that there are no braces after remove_all_X , because you don't want to pass the result of the function, but the function itself. Remember: filter() needs a function.

A function that is as simple as

def remove_all_X(s):
    return s != "X"

is often expressed as a lambda. You remove the definition ( def ) and write lambda + the parameter instead. The return keyword is also omitted. And you get:

labmda s: s != "X"

Now you use that instead of the function name:

message=filter(labmda s: s != "X" ,garbled)

What is wrong with my code?

Let's try moving backwards in the explanation, but starting from your code

lambda x: x=="X" x==''

Adding all the stuff that was removed:

def remove_all_X(x):
    return x=="X" x==''

Do you see that this does not result in a valid return statement? It looks like you want to return 2 things (maybe a tuple (x=="X", x=='') ) or a combination of two things (maybe x=="X" or x=="" )?

The compiler cannot interpret your wish and doesn't understand what to do.

Answer 3

It's because lambda and filter just don't work like that. This is what happens when you try the first one

>>> message=filter(lambda x: x=="X" x=='', garbled)
SyntaxError: invalid syntax

The second one basically allows every letter where letter!='X' (so where every letter is not 'X')

The Syntax Error appears because filter doesn't find every instance of a letter and replace it. It simply looks for a letter and excludes them, so putting the x=='' after the x=="X" doesn't make sense because there is an unexpected space, so Syntax Error ...