Stupid Lambda Tricks [LONG]

Gerson Kurz gerson.kurz at t-online.de
Thu Feb 21 10:51:39 CET 2002


Note: the official URL for the following is
http://www.p-nand-q.com/lambda.htm

-----------------------------------------------------
Stupid Lambda Tricks

This document describes some stupid tricks for using lambda
expressions in Python. It is not intended for the "functional
programming style" crowd, but for the "wacky geek fun" hackers.

To understand the need for such a page, the most common limitations on
lambda expressions are:

- You can only use expressions, but not statements. 
- You can only use one expression. 
- You cannot declare and use local variables. 

* Adding Statements to Lambda

The official python documentation says, that you can only use
expressions in lambda, but not statements. For example, the tutorial
says in section 4.7.4: "They (lambda expressions) are syntactically
restricted to a single expression". You can also see this from the
syntax definition in section 5.10 of the "Python Reference Manual":

expression:     or_test | lambda_form
or_test:        and_test | or_test "or" and_test
and_test:       not_test | and_test "and" not_test
not_test:       comparison | "not" not_test
lambda_form:    "lambda" [parameter_list]: expression

* Using print in lambda

The best solution is to define your own print function:

>>> out=lambda *x:sys.stdout.write(" ".join(map(str,x)))
>>> out("test",42)
test 42

Note that this function doesn't do string formatting - you can use the
normal %-style syntax for that.

* Using data assignment in lambda

In Python, assignments are statements (see "6.3 Assignment statements"
in the "Python Reference Manual"), so you cannot easily use them in
lambdas.

* Data assignment for lists

To set an item in a list, you can use the member function __setitem__.
For example, rather than writing

data[4] = 42

you can write

data.__setitem__(4,42)

Example: Here is a function that swaps two elements in a given list:

def swap(a,x,y):
    a[x] = (a[x], a[y])
    a[y] = a[x][0]
    a[x] = a[x][1]

You can write this as a lambda expression like this:

swap = lambda a,x,y:(lambda
f=a.__setitem__:(f(x,(a[x],a[y])),f(y,a[x][0]),f(x,a[x][1])))()

* Data assignment for locals

Sometimes you have local variables, and want to assign values to them.
The best solution is to store all local variables in a list, and use
list assignment as described above.

* Using IF statements

Here is a quote from section "4.16." from "The Whole Python FAQ:"

4.16. Is there an equivalent of C's "?:" ternary operator?
Not directly. In many cases you can mimic a?b:c with "a and b or c",
but there's a flaw: if b is zero (or empty, or None -- anything that
tests false) then c will be selected instead. In many cases you can
prove by looking at the code that this can't happen (e.g. because b is
a constant or has a type that can never be false), but in general this
can be a problem. Tim Peters (who wishes it was Steve Majewski)
suggested the following solution: (a and [b] or [c])[0]. Because [b]
is a singleton list it is never false, so the wrong path is never
taken; then applying [0] to the whole thing gets the b or c that you
really wanted. Ugly, but it gets you there in the rare cases where it
is really inconvenient to rewrite your code using 'if'. 

This method is of course trivially defined as lambda:

IF = lambda a,b,c:(a and [b] or [c])[0]

* Using WHILE statements

Here is a generic while loop:

while expression():
    function()

So, given expression and function, you can write this as:

WHILE = lambda e,f:(e() and (f(),WHILE(e,f)) or 0)

Which is sort-of tail recursive, but uses a lot of stack.

* Using FOR statements

Here is a generic for loop:

for x in items:
    f(x)

Use map() instead. You can rewrite this as:

map(f,items)

Which has the benefit of being faster in most cases, too.

* Using Multiple expressions in Lambda

There are several possible solutions to the problem of how to use
several distinct expressions in one lambda statement. The general form
of the problem is: How to use

    f1()
    f2()
    ...
    fn()

in one lambda statement. The basic idea used in most of the following
is to rewrite that as

for function in (f1, f2, f3):
    function()

* Use tuples and rely on evaluation order

If the functions are not related, you can say:

(f1(), f2(), ..., fn())

The current evaluation order is from left to right, so that should
work.

* Use tuples, and implement your own evaluation order

If you are a coward and fear that evaluation order will change in a
future release of python, you can use eval

map(eval,("f1()","f2()",...,"fn()"))

You can also use apply

map(lambda x:apply(x,()),(f1,f2,f3))

* Using local variables

The best idea is to define a helper lambda function that has the local
variables as argument. For example, say you have this function:

def f(x):
    result = range(20)
    return do_something(result,x)

You can rewrite this as

f = lambda x: (lambda result=range(20):do_something(result,x))()

As mentioned above, local variables are best used in a list, because
that way you can use [].__setitem__ for data assignments.

* Misc topics

Here are some other topics that don't seem to fit in any of the above.

* Wrappers for list functions that return None

Many list functions return None, so you cannot easily use them in a
stacked expression inside lambda. Examples of this are [].sort() and
[].reverse(). Instead of writing:

list.some_function_that_returns_None()
function(list)

you can write

function((list.some_function_that_returns_None(),list)[1])

Here is an example:

import random

sort = lambda x: (x.sort(),x)[1]
reverse  = lambda x: (x.reverse(),x)[1]

test = range(20)
print test
random.shuffle(test)
print sort(test)
print reverse(sort(test))





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