lambda xxx: foo(xxx) - is it the totally same as just foo?
used in construction like
map(lambda xxx: foo(xxx), my_things)
I dont see any difference:
def plus(a):
return a+1
for i in map(plus, [1,2,3]):
print(i)
for i in map(lambda a: plus(a), [1,2,3]):
print(i)
prints the same
2
3
4
No, lambda x: foo(x) is not the same as foo(x) but rather foo. The eta reduction is the conversion of lambda x: foo(x) to foo.
map(lambda a: plus(a), [1,2,3]) and map(plus, [1,2,3]) are related by replacing lambda a: plus(a) with plus an application of the eta reduction rule.
Yes, that's right. lambda x: foo(x) is the same as just calling foo(x).
Remember that a lambda is just an anonymous function that does something to x. In this case, it just calls foo on x and returns the result of foo(x).
For example, calling (lambda x: foo(x), 60) is similar to doing this:
def function2(y):
return y*2
def function1(x):
return function2(x)
function1(60)
Related
What's the pythonic way to combine a list of lambdas in a single function? For example:
lambdas = [lambda x, k=k: x+k for k in range(3)]
I would like to get this all in a single lambda similar to this but without having to type it out:
f = lambda x: lambdas[2](lambdas[1](lambdas[0](x)))
You can do this with functools.reduce like below:
from functools import reduce
lambdas = [lambda x, k=k: x+k for k in range(3)]
# x = 0
reduce(lambda x, l: l(x), lambdas, x)
# -> l[2](l[1](l[0](x)))
# step_1 : x = x , l = lambda[0] -> lambda[0](x)
# step_2 : x = lambda[0](x), l = lambda[1] -> lambda[1](lambda[0](x))
# step_3 : x = lambda[1](lambda[0](x)), l = lambda[2] -> lambda[2](lambda[1](lambda[0](x)))
The reduce function is defined to be exactly what you want. An alternative is to use a simple for loop.
def f(x):
for func in lambdas:
x = func(x)
return x
to do this with a lambda seems kind of weird.
Is there any specific reason why we cannot:
def function_chainer(lambdas):
def chained(x):
for function in lambdas:
x = function(x)
return chained
This solution is not a one-liner, but it is pythonic I believe.
If you really need a one-liner, you can use functools.reduce:
lambda x: functools.reduce(lambda a, f: f(a), lambdas, x)
The first argument to reduce governs the way of applying each subsequent element, the second is the iterable (here - our iterable of lambdas) and the last one is the initializer - the first value we want to pass to those lambda functions.
I have no clue how the code executes. This is a question on a practice test.
(lambda x: x(x))(lambda y:4)
The output is 4 but I don't know how the code runs. I think the steps are as follows:
lambda (x) is defined
executes lambda (x)
returns x(lambda(y))
lambda (y) is defined
returns 4
x(4)?
I know step 6 is wrong.
This is a tricky question.
Lambdas are anonymous functions. Let's define them to help you understand what is happening.
# (lambda x: x(x))
def func_1(func):
return func(func)
# (lambda y: 4)
def func_2(y):
return 4
>>> func_1(func_2) # Equivalent to (lambda x: x(x))(lambda y:4)
4
The second function is just a callable that returns the constant value of 4 regardless of the value of y, so func_2(10) and func_2(0) both return 4.
The first function is a callable with itself as the single argument to the function (x(x)). We just saw that the second function is a callable that will return the constant value 4 regardless of the input argument, so func_2(func_2) simply returns 4.
lambda expression works just like an unnamed function (anonymous functions). And works like this:
lambda (1): (2)
(1) are parameters
(2) the return
See:
def square(x):
return x**2
Using lambda expression:
square = lambda x: x**2
>>> square(2)
4
But think about the use, maybe you just want to use an function a few times, why store it? Create an anonymous function! How? running a lambda function:
>>> (lambda x: x**2)(2)
4
See? Not that hard. Let's think together.
(lambda x: x(x))(lambda y:4)
lambda y: 4 is passed to our lambda x function, now x = (lambda y: 4) (crazy thing!). x(x) are equals to (lambda y: 4)((lambda y: 4)). Look! We got an parameter to our first 'y' it is (lambda y: 4)! then the first function are executed, and returns 4. What about the second? It doesn't run, and don't need to run! if you do x(x(x)) the second function will run, but not the third one.
Don't think about "lambda is defined". A lambda is a value, like 4. You don't "define" a 4 - it just is. When evaluating a lambda, just replace its (evaluated) arguments into its body (adding parentheses where neccessary). Work it like a rewriting problem.
(lambda x: x(x))(lambda y:4)
# rewrite all `x` in `x(x)` to `lambda y:4` (two appearances):
(lambda y:4)(lambda y:4)
# rewrite all `y` in `4` to `lambda y:4` (zero appearances):
4
I'm writing a python script that should take a list of functions, written as lambda expressions, and return the compose of all the function, but, I have an arrow in the script, maybe because of the way I'm using lambda expression. It seems like even after I give the returned function a number value, I get back a function, and not a number. This is what I've wrote:
def compose1(lst):
if lst == []:
return lambda x: x
else:
temp = (lst[len(lst)-1])
for i in range(len(lst)-2,-1,-1):
temp = lambda x: lst[i](temp)
return lambda x: temp
this is a tast for the function I've wrote, which says I have a mistake.
f = compose1([lambda x: x+1, lambda x: x*2, lambda x: x-1])
for x in range(10):
assert (f(x) == 1 + (x - 1) * 2)
f = compose1([lambda x: x-1, lambda x: x*2, lambda x: x+1])
for x in range(10):
assert (f(x) == (x + 1) * 2) - 1
I would apreciate some halp on this problem..
thanks :)
def compose(*funcs):
"""
compose(func[,...[, func]]) -> function
Return the composition of functions.
For example, compose(foo, bar)(5) == foo(bar(5))
"""
if not all(callable(func) for func in funcs):
raise TypeError('argument must be callable')
funcs = funcs[::-1]
def composition(*args, **kwargs):
args = funcs[0](*args, **kwargs)
for func in funcs[1:]:
args = func(args)
return args
return composition
f = compose(*[lambda x: x+1, lambda x: x*2, lambda x: x-1])
for x in range(10):
assert f(x) == (1 + (x - 1) * 2)
f = compose(*[lambda x: x-1, lambda x: x*2, lambda x: x+1])
for x in range(10):
assert f(x) == ((x + 1) * 2) - 1
It looks like your loop is just reimplementing what reduce does. Here's a functional look at your function composition problem:
def compose1(fnlist):
if not fnlist:
return lambda x: x
# compose 1 function of x from two others
def compose2fns(fn1, fn2):
return lambda x : fn1(fn2(x))
# or if you really love lambdas
# compose2fns = lambda fn1,fn2: lambda x: fn1(fn2(x))
# use reduce to cumulatively apply compose2fns to the functions
# in the given list
return reduce(compose2fns, fnlist)
This passes your tests just fine.
CODE GOLF:
I couldn't resist, here is a one-liner, even including your check for an empty input list:
compose1 = lambda fnlist: reduce(lambda fn1,fn2: lambda x : fn1(fn2(x)),
fnlist or [lambda x:x])
Your issue here is your logic.
for i in range(len(lst)-2,-1,-1):
temp = lambda x: lst[i](temp)
return lambda x: temp
This will set temp to a function. lst is a list of functions, lst[i] is a function. You call that, to give a value, and then you make a new function with lambda. You then return a function that gives that function.
Your return value is a function, that gives a function, that gives a value, hence your issue.
As a note, this code has other problems. if lst == []: should be if not lst:, for example. You should also never iterate by index, and instead iterate by value, as Python is designed. I can't actually work out what you are trying to achieve with your code, which shows how hard to read iterating by index is.
Your code currently does this:
If there are no values, return a function that returns the first argument.
If there is one value, return the function in the list.
If there are many values, return a function that returns a function that returns the first value retrieved by running the first function from the list.
I'm not sure what you were trying to do, but I'm pretty sure that isn't it.
I like this syntax:
f = do (lambda x: x-1) (lambda x: x*2) (lambda x: x+1)
for x in range(10):
assert f(x) == 1 + (x - 1) * 2
The implementation is surprisingly trivial:
class do(object):
def __init__(self, x):
self.fns = [x]
def __call__(self, x):
if callable(x):
self.fns.append(x)
return self
for f in self.fns:
x = f(x)
return x
I'm having trouble with a question which follows: Write a recursive function repeatedlyApply that takes as arguments a function
f of one argument and a positive integer n. The result of repeatedlyApply is a function of one argument that applies f to that argument n times.
So, for example, we would have
repeatedlyApply(lambda x: x+1,10)(100) ==> 110
You may assume that the following function has been defined. You don't have to use it, but it can contribute to a pretty solution.
def compose(f,g):
return lambda x: f(g(x))
So far i've written this
def compose(f,g):
return lambda x: f(g(x))
def recApply(f,n):
for i in range(n):
return recApply(compose(f,f), n-1)
return f
I'm going wrong somewhere because using the above example recApply(lambda x: x+1,10)(100) i get 1124.
Help much appreciated
Correct answer is:
def recApply(func, n):
if n > 1:
rec_func = recApply(func, n - 1)
return lambda x: func(rec_func(x))
return func
And the output:
>>>> print recApply(lambda x: x+1,10)(100)
110
Your function needs some work:
You have a return inside your for loop, so you return immediately instead of running the loop.
You have a recursive call inside your for loop, so you are doing a bit too much iteration. Choose one or the other.
Be careful when you stack function compositions on top of each other, you are doing power composition rather than linear composition.
Can you tell us what precisely you are trying to do?
EDIT: Since everybody else is posting an answer:
recApply = lambda f, n: lambda x: x if n == 0 else recApply(f, n-1)(f(x))
I have a solution based on lambdas:
>>> f = lambda x: x + 10
>>> iterate = lambda f, n, x : reduce(lambda x, y: f(x), range(n), x)
>>> iterate(f, 10, 3)
103
>>> iterate(f, 4, 4)
44
>>> f10 = lambda x: iterate(f, 10, x)
>>> f10(5)
105
I assume this is an exercise of some sort. There are a few ways you could do it, here's a short one:
>>> repeatedlyApply = lambda f, n: reduce(lambda f1, f2: compose(f1, f2), [f]*n)
>>> repeatedlyApply(lambda x: x+1,10)(100)
110
Very rarely I'll come across some code in python that uses an anonymous function which returns an anonymous function...?
Unfortunately I can't find an example on hand, but it usually takes the form like this:
g = lambda x,c: x**c lambda c: c+1
Why would someone do this? Maybe you can give an example that makes sense (I'm not sure the one I made makes any sense).
Edit: Here's an example:
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])))()
You could use such a construct to do currying:
curry = lambda f, a: lambda x: f(a, x)
You might use it like:
>>> add = lambda x, y: x + y
>>> add5 = curry(add, 5)
>>> add5(3)
8
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])))()
See the () at the end? The inner lambda isn't returned, its called.
The function does the equivalent of
def swap(a, x, y):
a[x] = (a[x], a[y])
a[y] = a[x][0]
a[x] = a[x][1]
But let's suppose that we want to do this in a lambda. We cannot use assignments in a lambda. However, we can call __setitem__ for the same effect.
def swap(a, x, y):
a.__setitem__(x, (a[x], a[y]))
a.__setitem__(y, a[x][0])
a.__setitem__(x, a[x][1])
But for a lambda, we can only have one expression. But since these are function calls we can wrap them up in a tuple
def swap(a, x, y):
(a.__setitem__(x, (a[x], a[y])),
a.__setitem__(y, a[x][0]),
a.__setitem__(x, a[x][1]))
However, all those __setitem__'s are getting me down, so let's factor them out:
def swap(a, x, y):
f = a.__setitem__
(f(x, (a[x], a[y])),
f(y, a[x][0]),
f(x, a[x][1]))
Dagnamit, I can't get away with adding another assignment! I know let's abuse default parameters.
def swap(a, x, y):
def inner(f = a.__setitem__):
(f(x, (a[x], a[y])),
f(y, a[x][0]),
f(x, a[x][1]))
inner()
Ok let's switch over to lambdas:
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]))()
Which brings us back to the original expression (plus/minus typos)
All of this leads back to the question: Why?
The function should have been implemented as
def swap(a, x, y):
a[x],a[y] = a[y],a[x]
The original author went way out of his way to use a lambda rather then a function. It could be that he doesn't like nested function for some reason. I don't know. All I'll say is its bad code. (unless there is a mysterious justification for it.)
It can be useful for temporary placeholders. Suppose you have a decorator factory:
#call_logger(log_arguments=True, log_return=False)
def f(a, b):
pass
You can temporarily replace it with
call_logger = lambda *a, **kw: lambda f: f
It can also be useful if it indirectly returns a lambda:
import collections
collections.defaultdict(lambda: collections.defaultdict(lambda: collections.defaultdict(int)))
It's also useful for creating callable factories in the Python console.
And just because something is possible doesn't mean that you have to use it.
I did something like this just the other day to disable a test method in a unittest suite.
disable = lambda fn : lambda *args, **kwargs: None
#disable
test_method(self):
... test code that I wanted to disable ...
Easy to re-enable it later.
This can be used to pull out some common repetitive code (there are of course other ways to achieve this in python).
Maybe you're writing a a logger, and you need to prepend the level to the log string. You might write something like:
import sys
prefixer = lambda prefix: lambda message: sys.stderr.write(prefix + ":" + message + "\n")
log_error = prefixer("ERROR")
log_warning = prefixer("WARNING")
log_info = prefixer("INFO")
log_debug = prefixer("DEBUG")
log_info("An informative message")
log_error("Oh no, a fatal problem")
This program prints out
INFO:An informative message
ERROR:Oh no, a fatal problem
It is most oftenly - at least in code I come accross and that I myself write - used to "freeze" a variable with the value it has at the point the lambda function is created. Otherwise, nonlocals variable reference a variable in the scope they exist, which can lead to undesied results sometimes.
For example, if I want to create a list of ten functions, each one being a multiplier for a scalar from 0 to 9. One might be tempted to write it like this:
>>> a = [(lambda j: i * j) for i in range(10)]
>>> a[9](10)
90
Whoever, if you want to use any of the other factoried functions you get the same result:
>>> a[1](10)
90
That is because the "i" variable inside the lambda is not resolved when the lambda is created. Rather, Python keeps a reference to the "i" in the "for" statement - on the scope it was created (this reference is kept in the lambda function closure). When the lambda is executed, the variable is evaluated, and its value is the final one it had in that scope.
When one uses two nested lambdas like this:
>>> a = [(lambda k: (lambda j: k * j))(i) for i in range(10)]
The "i" variable is evaluated durint the execution of the "for" loop. Itś value is passed to "k" - and "k" is used as the non-local variable in the multiplier function we are factoring out. For each value of i, there will be a different instance of the enclosing lambda function, and a different value for the "k" variable.
So, it is possible to achieve the original intent :
>>> a = [(lambda k: (lambda j: k * j))(i) for i in range(10)]
>>> a[1](10)
10
>>> a[9](10)
90
>>>
It can be used to achieve a more continuation/trampolining style of programming,
See Continuation-passing style
Basically, with this you can modify functions instead of values
One example I stumbled with recently: To compute approximate derivatives (as functions) and use it as an input function in another place.
dx = 1/10**6
ddx = lambda f: lambda x: (f(x + dx) - f(x))/dx
f = lambda x: foo(x)
newton_method(func=ddx(f), x0=1, n=10)