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why is my recursion not returning the desired output? [closed]

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I am writing a recursion function that simply returns x^n:
def myPow(x,n) -> float:
def recurs_n(x,n):
if n==1:
return x
if n>1:
result = myPow(x,n-1)
return result
return recurs_n(x,n)
print(myPow(2,5))
The output that I am getting for the above is 2, which isn't correct obviously. Any idea where I am going wrong?

Your inner function calls the outer function, that should not be allowed. Furthermore, there is no logic applied anywhere for the multiplication when going to the n-1 step. So when n>1 you need to return x*myPow(x,n-1)
I should also mention, that in your case there is no real need to have an inner and outer function, one function should be enough. Like so;
def myPow(x,n) -> float:
if n == 1:
return x
if n > 1:
return x*myPow(x,n-1)
print(myPow(2,5))

Your code runs until power n=5 reaches 1, where it activates the first if condition where x is returned (here x=2) without continuing onwards.
An alternative is presented below:
def power(N, P):
# if power is 0 then return 1
if P == 0:
return 1
# if power is 1 then number is
# returned
elif P == 1:
return N
else:
return (N*power(N, P-1))

Can you find the even number problem using with try / except? [closed]

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I didn't understand this thing:
list1 = [34,2,1,3,33,100,61,1800]
for n in list1:
try:
n%2 == 0
print(n)
except:
pass
The output of the code above shows every number, but I need just even numbers.
What is my mistake?

I suspect you are missing a basic idea somewhere. try / except is used for catching errors. In order to use this in the way you are trying to, you need to cause an error under some condition. One easy way is to assert the number is even. assert n % 2 == 0 says if n is not even, raise an exception. Then you can catch the exception and skip it with a pass
list1 = [34,2,1,3,33,100,61,1800]
for n in list1:
try:
assert n % 2 == 0 # cause an error if `n` is not even
print(n)
except AssertionError:
pass

You were missing a condition if
list1 = [34,2,1,3,33,100,61,1800]
for n in list1:
try:
if n%2 == 0:
print(n)
except:
pass

Two things. You forgot an if statement before preforming n%2
Second thing is, what is the point of the try and except. You don't need it.
Below is an example of your code in
list comprehension.
list1 = [34,2,1,3,33,100,61,1800]
even = [n for n in list1 if n%2 == 0]
output
[34, 2, 100, 1800]
simplified code
list1 = [34,2,1,3,33,100,61,1800]
even = []
for n in list1:
if n%2 == 0:
even.append(n)
output
[34, 2, 100, 1800]

basic question about python for loop and enumerate [closed]

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I am trying to build a function that will return the sum of the first item in the data list and every tenth item after, so I write the script below, however, the result is always the first data in my list. How should I modify my code and fix it?
def Sum10th(data):
sum=0
for i,d in enumerate(data):
if (i % 10 == 0): sum = sum+d
return sum
p = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]
print(Sum10th(p))
The desired result should be 31, however the computer only returns the value of 1.

You can try
p = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]
def Sum10th(data):
return sum([v for i, v in enumerate(p) if (i + 1) % 10 == 0 or i == 0])
print(Sum10th(p))
Output
31
By using the list comprehension with an if, at the end of it, you can get all the first item with the tenth item following it from a list.

Try this way:
def sum10thdata(data):
sum = 0
l = len(data)
for i in range(0,l,10):
sum+=data[i]
return sum

Appending an array to an array using a function in python [closed]

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I'm trying to append an array to an array using a function but it seems to return None.
Code (now works):
import random
arr = []
def randomarr():
t = []
for i in range(3):
r = random.randint(1,9)
t.append(r)
return t
for i in range(3):
arr.append(randomarr())
print(matrix)
Edit: Solved by adding a return. Shame on me for forgetting to put a return statement in my code.

Your funtion randomarr needs a return staterment.
try:
def randomarr():
t = []
for i in range(3):
r = random.randint(1,9)
t.append(r)
return t

randomarr() is not retuning anything, is just printing t.
use return like this:
import random
arr = []
def randomarr():
t = []
for i in range(3):
r = random.randint(1,9)
t.append(r)
return t
for i in range(3):
arr.append(randomarr())
print(arr)

Trying to average a list of numbers using functions [closed]

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First and foremost, I am new to python. As of such, I only know for loops, range, lens, and sum to do this problem. I am having difficulty trying to make a function that can average list of numbers.
This is my code so far:
def ave(L):
L = list(range(len(L))
for a in range(len(L)):
if len(L) == 0: return 0
else: return float((sum(L))/len(L))
So far, I am getting a syntax error on my third line with range(L).

All you need to do is return the sum of L divided by the length of L:
def ave(L):
if not L:
return 0
return sum(L) / len(L)
No range() or float() or for required.
In Python 3, / always produces a floating point number. sum() will do all the looping for you. The only thing you need to take care of, is returning 0 if the list is empty.

Following line is missing a ):
L = list(range(len(L)))
^

Because average of empty list is undefined, you should rather return None instead of '0'.
And instead checking for length, it is better to catch potential error, according to EAFP principle. It makes also more clear what are you doing, as error is self-descriptive.
def ave(L):
try:
return sum(L) / len(L)
except ZeroDivisionError:
return None