deeps deeps - 1 year ago 103
Python Question

lambda in python with extra unpassed variable

I am new to python scripting.
I am confused how lambda interprets variables passed as in below example.

def create_multipliers():
return [lambda x : i * x for i in range(5)]

for multiplier in create_multipliers():
print multiplier(2),

returns 8 8 8 8 8

I see that lambda accepts only one argument (i.e 'x').

How does it interpret x and i in create_multipliers?
Also what does multiplier(2) mean?

Please help

Also with the below example

def make_incrementor (n): return lambda x: x + n
print make_incrementor(22)(33)

returned 55

How does the lambda/make_incrementor function decide which is 'x' and 'n'?

Answer Source

The first part of code creates a list of lambdas which each take single argument x and multiplies it with i. Note that each lambda is bound to variable i not its' current value. Since the value of i after the list comprehension is 4 each lambda will return x * 4:

>>> def create_multipliers():
...     return [lambda x: i * x for i in range(5)]
>>> l = create_multipliers()
>>> l[0](1)
>>> l[4](1)
>>> l[4]('foobar')

The loop will then execute each lambda with parameter 2 and print the results to same line. Since there are 5 lambdas and 4 * 2 is 8 you get the output that you see. , after the print statement will result the output to be printed on the same line:

>>> for multiplier in l:
...     print multiplier(2),
8 8 8 8 8

make_incrementor works with same principle. It returns a lambda that takes single argument x which is "decided" when lambda is called. It will then return x + n where n is the value passed to make_incrementor:

>>> def make_incrementor(n):
...     return lambda x: x + n
>>> inc = make_incrementor(2) # n is "decided" here
>>> inc(3)                    # and x here

UPDATE Example with nested functions:

>>> def make_incrementor(n):
...     def nested(x):
...         return x + n
...     return nested
>>> inc = make_incrementor(2)
>>> inc(3)
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