Python lambda if statement re.sub

Question

So I am using the following regex to parse text and grab information from a specific dictionary:

re.sub(r'(<Q\d+>)',lambda m: quotes[m.group(1)][1],text)

What I want to do, is only have it replace if what it would replace with is a key in a separate dictionary. Logically it would look like this:

re.sub(r'(<Q\d+>)',lambda m: quotes[m.group(1)][1] if quotes[m.group(1)][1] in d,text)

now if I were to run the following, I get the following syntax error:

>>> re.sub(r'(<Q\d+>)',lambda m: quotes[m.group(1)][1] if quotes[m.group(1)][1] in d,text)
  File "<stdin>", line 1
    re.sub(r'(<Q\d+>)',lambda m: quotes[m.group(1)][1] if quotes[m.group(1)][1] in d,text)
                                                                                    ^
SyntaxError: invalid syntax

How can I only replace in this way?

Bostone · Accepted Answer · 2019-02-04 21:54:37Z

12

The if expression always requires an else. You always have to replace the matched text. If you don't want to replace it, you just need to replace it with itself:

re.sub(r'(<Q\d+>)', 
  (lambda m: quotes[m.group(1)][1] if quotes[m.group(1)][1] in d else m.group(1)), text)

edited Feb 4, 2019 at 21:54

Bostone

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answered Sep 20, 2013 at 17:45

BrenBarn

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4 Comments

Ryan Saxe Over a year ago

same syntax error...also why would the else statement go after the comma that seperates the replace statement from the text?

BrenBarn Over a year ago

@RyanSaxe: Sorry, I made a typo and left the text in the wrong place, see the updated version. (You should put spaces after your commas! :-)

Ryan Saxe Over a year ago

are the parentheses wrapping the lambda necessary?

abarnert Over a year ago

@RyanSaxe: The parens are not necessary here, because it's not ambiguous to the parser whether the comma is inside the lambda body or the lambda is a comma-separated argument to re.sub. However, if it's not immediately obvious to you that it's not ambiguous, you may want to use them anyway, because code has to be meaningful to human readers, not just to the compiler.

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