I have a python function which takes as input an arbitrary number of parameters. I beleive this is implemented via *args (thanks #Anurag Wagh for pointing this out). I would like to be able to pass to it a list of values. Here is the specific code I have:
from sympy import solve_poly_system, symbols
x,y,z = symbols('x y z')
print(solve_poly_system([y**2 - x**3 + 1, y*x], x, y))
Instead of passing x and y to solve_poly_system, I want to be able to pass it a list L something like so:
print(solve_poly_system([y**2 - x**3 + 1, y*x], L))
where L = [x,y].
Is this possible? Please note, although I am interested in the specific case of the solve_poly_system function, I am more broadly interested in the passing a list to any function in python that takes an arbitrary number of inputs.
To put this another way, I want to know how to pass a list of values to a function which takes *args as input. In particular I am interested in the case where the length of the list is not known until run time and may change during run time.
From your question, it looks like you know how to use *args in function definition and you are looking for a way to pass all the elements of a list to that function regardless of the length of the list. If that's your question, then here's your answer:
def function(x, *args):
print(x)
print(args)
x = 10
l = [5, 6, 7, 8]
function(x, *l)
Output:
10
(5, 6, 7, 8)
Related
I wrote the following function
def addsub(a):
def add(a):
def subtract(b):
return a-b
return subtract
return add(a)
addsub(9)(4)
returns 5
but what if I do not know to number of add subtract I want to perform
addsub(9)(3)(4)(5).
The above function does not work for it
could I write something general which works for any length of input?
PS. I do not want to use functools
It is not possible exactly like you describe (with currying). But if all you want is to sum the numbers of a list with a varying sign, this looks the most straightforward to me:
import itertools
coefficients = itertools.cycle((1, -1))
numbers = [9, 3, 4, 5]
result = sum(a * c for a, c in zip(numbers, coefficients))
I have this function that I need to compare the strings in a list to a *args
The reason being is that, the user should be able to type any words in the 2nd argument. However when I try to compare the strings to the *args it doesn't give me any results
def title_case2(title, *minor_words):
for x in title.split():
if x in minor_words:
print(x)
Assuming I ran the function with the parameters below. I was hoping it would display a and of since these words are found on those 2 entries.
title_case2('a clash of KINGS','a an the of')
*args is a tuple of arguments, so you're actually checking if x is in ('a an the of',). So either pass your argument as:
title_case2('a clash of KINGS', *'a an the of'.split())
Or, use this as your test:
if any(x in y for y in minor_words):
In either of the above cases the output is:
a
of
This is one approach.
Ex:
def title_case2(title, *minor_words):
minor_words = [j for i in minor_words for j in i.split()] #Create a flat list.
for x in title.split():
if x in minor_words:
print(x)
title_case2('a clash of KINGS','a an the of', "Jam")
using a for-loop instead of list comprehension
def title_case2(title, *minor_words):
minor_words_r = []
for i in minor_words:
for j in i.split():
minor_words_r.append(j)
for x in title.split():
if x in minor_words_r:
print(x)
So, Python functions can return multiple values. It struck me that it would be convenient (though a bit less readable) if the following were possible.
a = [[1,2],[3,4]]
def cord():
return 1, 1
def printa(y,x):
print a[y][x]
printa(cord())
...but it's not. I'm aware that you can do the same thing by dumping both return values into temporary variables, but it doesn't seem as elegant. I could also rewrite the last line as "printa(cord()[0], cord()[1])", but that would execute cord() twice.
Is there an elegant, efficient way to do this? Or should I just see that quote about premature optimization and forget about this?
printa(*cord())
The * here is an argument expansion operator... well I forget what it's technically called, but in this context it takes a list or tuple and expands it out so the function sees each list/tuple element as a separate argument.
It's basically the reverse of the * you might use to capture all non-keyword arguments in a function definition:
def fn(*args):
# args is now a tuple of the non-keyworded arguments
print args
fn(1, 2, 3, 4, 5)
prints (1, 2, 3, 4, 5)
fn(*[1, 2, 3, 4, 5])
does the same.
Try this:
>>> def cord():
... return (1, 1)
...
>>> def printa(y, x):
... print a[y][x]
...
>>> a=[[1,2],[3,4]]
>>> printa(*cord())
4
The star basically says "use the elements of this collection as positional arguments." You can do the same with a dict for keyword arguments using two stars:
>>> a = {'a' : 2, 'b' : 3}
>>> def foo(a, b):
... print a, b
...
>>> foo(**a)
2 3
Actually, Python doesn't really return multiple values, it returns one value which can be multiple values packed into a tuple. Which means that you need to "unpack" the returned value in order to have multiples.
A statement like
x,y = cord()
does that, but directly using the return value as you did in
printa(cord())
doesn't, that's why you need to use the asterisk. Perhaps a nice term for it might be "implicit tuple unpacking" or "tuple unpacking without assignment".
Given the following:
(I) a = map(function, sequence)
(II) a = [function(x) for x in sequence]
When would I need to use (I)? Why choose a map object over a list when the latter is subscriptable and IMO more readable?
Also, could someone explain line 6 of the following code (Python 3):
>>>import math
>>>a = map(int,str(math.factorial(100)))
>>>sum(a)
648
>>>sum(a)
0
Why is the sum of the map object changing?
When would I need to use (I)? Why choose a map object over a list when the latter is subscriptable and IMO more readable?
map was introduced in Python 1.0, while list comprehension was not introduced until Python 2.0.
For Python 2+, you never need to use one or the other.
Reasons for still using map could include:
preference. You prefer list comprehension, not everyone agrees.
familiarity. map is very common across languages. If Python's not your native language, "map" is the function you'll look up.
brevity. map is often shorter. Compare map and lambda f,l: [f(x) for x in l].
I is an iterator -- it creates a stream of values which then vanish. II is a list -- it lasts for a while and has lots of features, like len(mylist) and mylist[-3:].
The sum changes because the iterator vanishes after you use it.
Use lists and list comprehensions. If you process tons of data, then iterators (and generators, and generator comprehensions) are awesome, but they can be confusing.
Or, use an iterator and convert into a list for further processing:
a = list( map(int,str(math.factorial(100))) )
From the docs:
Apply function to every item of iterable and return a list of the results. If additional iterable arguments are passed, function must take that many arguments and is applied to the items from all iterables in parallel...
The sum changes to 0 because the iterator is iterated, so it becomes nothing. This is the same concept with .read() (Try calling x = open('myfile.txt'), and then type print x.read() twice.)
In order to preserve the iterable, surround it with list():
>>> import math
>>> a = map(int,str(math.factorial(100)))
>>> sum(a)
648
>>> sum(a)
0
>>> a = list(map(int,str(math.factorial(100))))
>>> sum(a)
648
>>> sum(a)
648
Example from the docs:
>>> seq = range(8)
>>> def add(x, y): return x+y
...
>>> map(add, seq, seq)
[0, 2, 4, 6, 8, 10, 12, 14]
I am trying to understand these instructions.
Set up a new function in your main program file named “summer” that takes a list as a parameter and returns a value we will determine in the next steps.
In the “summer” function, set up a loop that uses a counter variable named “n” that will take on the values 0, 2, 4, 6, 8, 10, 12.
Each time through the loop, you are to call your “powerval” function from the “mymath” module passing as parameters item “n” and “n+1” from the list of data passed into “summer”. Add up all these values and return the final result to the caller.
So far I have:
def summer(list):
for n in range(0,13,2):
value=powerval(n,n+1)
After that I am lost. How do i perform step 3?
You add them up:
from mymath import powerval
def summer(somelist):
sum = 0
for n in range(0, 13, 2):
sum += powerval(somelist[n], somelist[n + 1])
return sum
So the return value of powerval() is added to the total sum so far, which was started at 0. You do need to pass in the somelist[n] and somelist[n + 1] values, not the indices themselves.
You need to add them up:
from mymath import powerval
def summer(lst):
total = 0
for n in range(0, 13, 2):
total += powerval(lst[n], lst[n + 1])
return total
I'm not sure where you use lst (I renamed list to lst, as list is a built-in function), so I'm guessing you're trying to get the nth and n + 1th elements from that list.
You can use the sum method to accomplish this in a very fashionable way :)
def summer(myList):
return sum(powerval(myList[n], myList[n+1]) for n in range(0, 13, 2))
This is also the fastest way.
PS: It's not a good idea to name you list "list", bacause that's a reserved name in python. That's why I have renamed it to myList in the example above.