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Why doesn't print work in a lambda?

Posted by: admin November 1, 2017 Leave a comment

Questions:

Why doesn’t this work?

lambda: print "x"

Is this not a single statement, or is it something else?
The documentation seems a little sparse on what is allowed in a lambda…

Answers:

A lambda‘s body has to be a single expression. In Python 2.x, print is a statement. However, in Python 3, print is a function (and a function application is an expression, so it will work in a lambda). You can (and should, for forward compatibility 🙂 use the back-ported print function if you are using the latest Python 2.x:

In [1324]: from __future__ import print_function

In [1325]: f = lambda x: print(x)

In [1326]: f("HI")
HI

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Answers:

what you’ve written is equivalent to

def anon():
    return print "x"

which also results in a SyntaxError, python doesn’t let you assign a value to print in 2.xx; in python3 you could say

lambda: print('hi')

and it would work because they’ve changed print to be a function instead of a statement.

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In cases where I am using this for simple stubbing out I use this:

fn = lambda x: sys.stdout.write(str(x) + "\n")

which works perfectly.

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The body of a lambda has to be an expression that returns a value. print, being a statement, doesn’t return anything, not even None. Similarly, you can’t assign the result of print to a variable:

>>> x = print "hello"
  File "<stdin>", line 1
    x = print "hello"
            ^
SyntaxError: invalid syntax

You also can’t put a variable assignment in a lambda, since assignments are statements:

>>> lambda y: (x = y)
  File "<stdin>", line 1
    lambda y: (x = y)
                 ^
SyntaxError: invalid syntax

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You can do something like this.

Create a function to transform print statement into a function:

def printf(text):
   print text

And print it:

lambda: printf("Testing")

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The body of a lambda has to be a single expression. print is a statement, so it’s out, unfortunately.

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Here, you see an answer for your question. print is not expression in Python, it says.

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With Python 3.x, print CAN work in a lambda, without changing the semantics of the lambda.

Used in a special way this is very handy for debugging.
I post this ‘late answer’, because it’s a practical trick that I often use.

Suppose your ‘uninstrumented’ lambda is:

lambda: 4

Then your ‘instrumented’ lambda is:

lambda: (print (3), 4) [1]