I have a helping function:
def incr(x):
return x+1
I want to create a function named "repeated" that use "incr" function n times on a certain parameter
In the end I want to use the "repeated" function in this matter only :
repeated (incr, 4)(2)
That for example will output 6.
So far I tried to do this:
def repeated(f, n):
func, x = f
for i in range(n):
func(x)
But it gave me an error saying I can't unpack a non Tuple function.
It doesn't seem like I don't have access in the function to the "(2)"
I do not recommend to use such a syntax construct, for such a task:
repeated(incr, 4)(2)
Your repeated function must return another function, that will be called by (2).
This should work in your requested manner:
def incr(x):
return x+1
def repeated(f, x):
# function foo will be returned by repeated and called by (2)
def foo(n):
res = x
for i in range(n):
res = f(res)
return res
return foo
print(repeated(incr, 4)(2))
I think you may want to do something like functional programming.
Add args to deal with for different kind of function you want to repeat.
I can't confirm if there is a position argument what kind of results you want, so I didn't deal with it.
code:
import functools
def incr(x):
return x + 1
def incx(x,y = 0):
return x + y + 1
def repeated_inner(func,args,times):
head, *rest = args
for _ in range(times):
head = func(head, *rest)
return args[0]
def repeated(func, *args ):
return functools.partial(repeated_inner, func, args)
print(repeated(incr, 4)(2))
print(repeated(incx, 4)(2))
print(repeated(incx, 4 ,3)(2))
result
6
6
12
the repeatedfunction must return a function
def repeated(func, n):
def repeatedfunc(x):
rsl = x
for i in range(n):
rsl = func(rsl)
return rsl
return repeatedfunc
def incr(x):
return x+1
rslt = repeated(incr, 4)(2)
print(rslt)
output
6
You should write something like this:
def repeated(f, arg_0, n):
arg = arg_0
for i in range(n):
arg = f(arg)
return arg
In a more general situation:
def repeated(f, arg):
def n_f(n):
result = 0
for i in range(n):
result =f(arg)
return result
return n_f
Related
Say that I have a list of functions: [f1, f2, f3] (in python)
How do I return a single function F:=f3(f2(f1())) (notice that F is of function type). I know that it's possible to do it with .reduce() but I was wondering if there's another way to do it without using libraries.
edit:
Also, notice that the length of the list can be greater than 3
I tried:
def func(list):
i = 1
new_function = filters[0]
while i<=len(filters)-1:
new_function = filters[i](new_function)
i+=1
return new_function
but it doesn't work
The problem in your code is that you pass a function as argument with filters[i](new_function).
I would suggest this recursive solution:
def chain(first, *rest):
return lambda x: first(chain(*rest)(x) if rest else x)
Example use:
def inc(x):
return x + 1
def double(x):
return x * 2
def square(x):
return x * x
f = chain(square, double, inc)
print(f(5)) # ((5+1)*2) ** 2 == 144
I see that in the code you tried, you never actually call the first of your functions. (I also assume that your code starts: def func(filters):
Taking into account that f1() takes no parameter, but the rest take the parameter of the return of the previous function, this should work:
def fit(funcs):
v = funcs[0]()
for f in funcs[1:]:
v = f(v)
return v
def f1():
return 42
def f2(x):
return x
def f3(x):
return x
fs = [f1, f2, f3]
a = lambda:fit(fs)
print(a())
Output: 42
def get_single_func(func_list):
def single_func(*args, **kwargs):
ret = func_list[0](*args, **kwargs)
for func in func_list[1:]:
ret = func(ret)
return ret
return single_func
add_digits2(1)(3)(5)(6)(0) should add up all the numbers and stop when it reaches 0.
The output should be 15
The below code works but uses a global variable.
total = 0
def add_digits2(num):
global total
if num == 0:
print(total)
else:
total += num
return add_digits2
The result is correct but needs to do the same thing without using the global variable.
One thing you could do is use partial:
from functools import partial
def add_digits2(num, total=0):
if num == 0:
print(total)
return
else:
total += num
return partial(add_digits2, total=total)
add_digits2(2)(4)(0)
You can just pass in *args as a parameter and return the sum
def add_digits2(*args):
return sum(args)
add_digits2(1, 3, 5 ,6)
You could also use a class, using the __call__ method to obtain this behavior:
class Add_digits:
def __init__(self):
self.total = 0
def __call__(self, val):
if val != 0:
self.total += val
return self
else:
print(self.total)
self.total = 0
add_digits = Add_digits()
add_digits(4)(4)(0)
# 8
add_digits(4)(6)(0)
# 10
though I still don't get why you would want to do this...
Really hard to say what they are after when asking questions like that but the total could be stored in a function attribute. Something like this
>>> def f():
... f.a = 3
>>> f()
>>> f.a
3
I have been asked to find the modulo 37 (%37) of a function.
Ensure that your function returns an integer between 0 and 36.
Assume that all arguments are integers between 0 and 36.
Write a decorator #normalize_37 that does just that. I.e. if 'bar' is a function, then the decorated function will have all its arguments reduced modulo 37 and its return value reduced modulo 37.
Find:
#normalize_37
def add(x,y):
return x+y
print(add(45,67))
#where the answer is 1.
#normalize_37
def bar(n):
if n >= 37 or n <= -1:
raise ValueError
else:
return n
print(bar(123))
#where the answer is 12
So far I have came up with this for the first bit from looking up decorators online:
import math
def document(f):
def wrap(x,y):
print("I am going to find modulo 37 of", x,y)
f(x,y)
return wrap
#document
def add(x,y):
print(add(x,y)%37)
add(45,67)
but it isn't working for me and when I run it, it just repeats the "I am going to find modulo 37 of" bit over and over again.
The specific form requested would be
def normalize_37(fn): # the decorator
def inner(x, y): # the new decorated function
return fn(x, y) % 37
return inner
#normalize_37
def add(x, y):
return x + y
add(45, 67) # returns 112 % 37 == 1
but we can generalize one step further,
def normalize(n): # make a decorator!
def decorator(fn): # the decorator
def inner(*args): # the new decorated function
return fn(*args) % n
return inner
return decorator
#normalize(37)
def add(*args):
return sum(args)
add(45, 67) # returns 112 % 37 == 1
Edit: ok, I missed the bit about reducing the arguments as well. So it becomes
def normalize(n): # make a decorator!
def decorator(fn): # the decorator
def inner(*args): # the new decorated function
return fn(*(arg % n for arg in args)) % n
return inner
return decorator
#normalize(37)
def add(*args):
return sum(args)
add(45, 67) # returns ((45 % 37) + (67 % 37)) % 37 == 1
or in the simpler form,
def normalize_37(fn): # the decorator
def inner(x, y): # the new decorated function
return fn(x % 37, y % 37) % 37
return inner
I have a list which I want to sort by multiple keys, like:
L = [ ... ]
L.sort(key = lambda x: ( f(x), g(x) ))
This works fine. However, this results with unnecessary calls to g, which I would like to avoid (for being potentially slow). In other words, I want to partially and lazily evaluate the key.
For example, if f is unique over L (i.e. len(L) == len(set(map(f,L)))) no calls to g should be made.
What would be the most elegant/pythonic way to do this?
One way I can think of is to define a custom cmp function (L.sort(cmp=partial_cmp)), but IMO this is less elegant and more complicated than using the key parameter.
Another way would be to define a key-wrapper class which takes a generator expression to generate the different parts of the key, and override the comparison operators to compare one-by-one. However, I'm feeling there must be a simpler way...
EDIT: I'm interested in a solution for the general problem of sorting by multiple functions, not only two as in my example above.
You can try using itertools.groupby:
result = []
for groupKey, group in groupby(sorted(L, key=f), key=f):
sublist = [y for y in group]
if len(sublist) > 1:
result += sorted(sublist, key=g)
else:
result += sublist
Another possibility, even less elegant, but in place:
L.sort(key = f)
start = None
end = None
for i,x in enumerate(L):
if start == None:
start = i
elif f(x) == f(L[start]):
end = i
elif end == None:
start = i
else:
L[start:end+1] = sorted(L[start:end+1], key=g)
start = None
if start != None and end != None:
L[start:end+1] = sorted(L[start:end+1], key=g)
First version generalized to any number of functions:
def sortBy(l, keyChain):
if not keyChain:
return l
result = []
f = keyChain[0]
for groupKey, group in groupby(sorted(l, key=f), key=f):
sublist = [y for y in group]
if len(sublist) > 1:
result += sortBy(sublist, keyChain[1:])
else:
result += sublist
return result
The second version generalized to any number of functions (not fully in place though):
def subSort(l, start, end, keyChain):
part = l[start:end+1]
sortBy(part, keyChain[1:])
l[start:end+1] = part
def sortBy(l, keyChain):
if not keyChain:
return
f = keyChain[0]
l.sort(key = f)
start = None
end = None
for i,x in enumerate(l):
if start == None:
start = i
elif f(x) == f(l[start]):
end = i
elif end == None:
start = i
else:
subSort(l, start, end, keyChain)
start = i
end = None
if start != None and end != None:
subSort(l, start, end, keyChain)
Given a function, you could create a LazyComparer class like this:
def lazy_func(func):
class LazyComparer(object):
def __init__(self, x):
self.x = x
def __lt__(self, other):
return func(self.x) < func(other.x)
def __eq__(self, other):
return func(self.x) == func(other.x)
return lambda x: LazyComparer(x)
To make a lazy key function out of multiple functions, you could create a utility function:
def make_lazy(*funcs):
def wrapper(x):
return [lazy_func(f)(x) for f in funcs]
return wrapper
And together they could be used like this:
def countcalls(f):
"Decorator that makes the function count calls to it."
def _f(*args, **kwargs):
_f._count += 1
return f(*args, **kwargs)
_f._count = 0
return _f
#countcalls
def g(x): return x
#countcalls
def f1(x): return 0
#countcalls
def f2(x): return x
def report_calls(*funcs):
print(' | '.join(['{} calls to {}'.format(f._count, f.func_name)
for f in funcs]))
L = range(10)[::-1]
L.sort(key=make_lazy(f1, g))
report_calls(f1, g)
g._count = 0
L.sort(key=make_lazy(f2, g))
report_calls(f2, g)
which yields
18 calls to f1 | 36 calls to g
36 calls to f2 | 0 calls to g
The #countcalls decorator above is used to connfirm that when f1 returns a lot
of ties, g is called to break the ties, but when f2 returns distinct values,
g does not get called.
NPE's solution adds memoization within the Key class. With the solution above,
you could add memoization outside (independent of) the LazyComparer class:
def memo(f):
# Author: Peter Norvig
"""Decorator that caches the return value for each call to f(args).
Then when called again with same args, we can just look it up."""
cache = {}
def _f(*args):
try:
return cache[args]
except KeyError:
cache[args] = result = f(*args)
return result
except TypeError:
# some element of args can't be a dict key
return f(*args)
_f.cache = cache
return _f
L.sort(key=make_lazy(memo(f1), memo(g)))
report_calls(f1, g)
which results in fewer calls to g:
10 calls to f1 | 10 calls to g
You could use a key object that would lazily evaluate and cache g(x):
class Key(object):
def __init__(self, obj):
self.obj = obj
self.f = f(obj)
#property
def g(self):
if not hasattr(self, "_g"):
self._g = g(self.obj)
return self._g
def __cmp__(self, rhs):
return cmp(self.f, rhs.f) or cmp(self.g, rhs.g)
Here is an example of use:
def f(x):
f.count += 1
return x // 2
f.count = 0
def g(x):
g.count += 1
return x
g.count = 0
L = [1, 10, 2, 33, 45, 90, 3, 6, 1000, 1]
print sorted(L, key=Key)
print f.count, g.count
i want to make a decoration method to assign the variable which the function would use but wouldn't be deliver by itself.
for example add new variable y in lambda r,i wrote code in this way but didn't work.
r = lambda x:x+y
def foo(func):
def wrapped(*args,**kwargs):
y = 3
return func(y=y,*args,**kwargs)
return wrapped
r = foo(r)
print(r(444))
this wouldn't work too
r = lambda x:x+y
def foo(func):
def wrapped(*args,**kwargs):
y = 3
return func(*args,**kwargs)
return wrapped
r = foo(r)
print(r(444))
kwargs is a casual python dict type, so you can just set the value of the y key to be 3
r = lambda x, y=0:x+y
def foo(func):
def wrapped(*args,**kwargs):
print type(kwargs) # will output <type 'dict'>
kwargs['y'] = 3
return func(*args,**kwargs)
return wrapped
In Understanding kwargs in Python this is explained in details.
Problem is that the function r can accept only one argument, you need to change its definition to accept more args:
r = lambda x, y=0, *args, **kwargs: x + y
def foo(func):
def wrapped(*args,**kwargs):
y = 3
return func(y=y, *args,**kwargs)
return wrapped
r = foo(r)
print(r(444))
#447