Reputation: 31
Am new to python and I have been trying to solve this problem but it does not seem to work as intended. your help is highly appreciated:
Given two numbers X and Y, write a function that:
.
def number(x,y):
if x > y:
for i in range(x,y):
if i%2 == 0:
list = []
return list.append[i]
else:
for i in range(x,y):
if i%2 == 1:
list = []
return list.append[i]
print(number(10,2))
Upvotes: 2
Views: 1981
Reputation: 19
Here in this i use the list comprehensions.list comprehension is a easy and readable technique in python.In this i include both x and y
def fun(x,y):
if x>y:
l=[i for i in range(y,x-1) if i%2==0]
return l.reverse()
else:
l=[i for i in range(x,y+1) if i%2!=0]
return l
Upvotes: 0
Reputation: 4606
Knowing that 2 % 2 == 0
we then can just use if not 2 % 2
for evens since not 0
will evaluate to true, here it is with comprehension and in extended form
def something(x, y):
if x > y:
l = [i for i in range(y, x) if not i % 2]
else:
l = [i for i in range(x, y) if i % 2]
return l
print(something(10, 2))
print(something(2, 10))
~/python/stack$ python3.7 sum.py [2, 4, 6, 8] [3, 5, 7, 9]
Full loop:
def something(x, y):
l = []
if x > y:
for i in range(y, x):
if not i % 2:
l.append(i)
else:
for i in range(x, y):
if i %2:
l.append(i)
return l
Upvotes: 0
Reputation: 706
It's so easy to do, and there are several ways to do what do you want, so i show you two ways to do that, first an understandable way and second an easy way ok let's start:-
First example
def number(x,y):
list = [] #firstly create a list
if x > y: #if x was greater than y
for num in range(y, x): # a loop for searching between them
if(num % 2 == 0): # if the number was even add it to list
list.append(num)
elif y > x: #if y was greater than x
for num in range(x, y): # a loop for searching between them
if(num % 2 != 0): # if the number was not even add it to list
list.append(num)
return list
print(number(10, 20))
print(number(20, 10))
#[11, 13, 15, 17, 19]
#[10, 12, 14, 16, 18]
Second example
number = lambda x, y : [n for n in range(y, x) if n%2 == 0] if x > y else [n for n in range(x, y) if n%2 != 0]
print(number(10, 20))
print(number(20, 10))
#[11, 13, 15, 17, 19]
#[10, 12, 14, 16, 18]
Note : But be sure that in both of my answers the x
number is inclusive(exists in searching function) and the y
number is exclusive, so if you wanted to make both of them inclusive so make loops ...(x, y+1)...
and if you wanted to make both of them exclusive just change loops to ...(x+1, y)...
.
Upvotes: 0
Reputation: 51623
Instead of testing for oddness/evenness all the time, use range(start,stop[,step]) with a step of 2 starting with a (corrected, known) odd/even number:
def number(x,y):
if x > y:
if y%2 == 1: # y is smaller && odd
y += 1 # make even
return list(range(y,x,2)) # x is > y - start from y to x
else: # this is strictly not needed - but more verbose intention-wise
if x%2 == 0: # is even
x += 1 # make odd
return list(range(x,y,2))
print(number(10,32))
print(number(10,2))
x
and y
if x > y
list
iteratively - simply stuff the range-sequence into the list(sequence)
constructor and return itOutput:
[11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31]
[2, 4, 6, 8]
Upvotes: 1
Reputation: 12305
And it is done. Basically if x > y
, you need to switch the first range
. You append the items normally(using ()
instead of []
), and then return the full list, got it?
def number(x,y):
list = []
if x > y:
for i in range(y,x):
if i%2 == 0:
list.append(i)
else:
for i in range(x,y):
if i%2 == 1:
list.append(i)
return list
print(number(10,2))
Working sample: https://py3.codeskulptor.org/#user302_nwBq00w56n_1.py
Upvotes: 1
Reputation: 896
Try this code it's working as per your need.
def number(x,y):
num= []
if x > y:
for i in range(y,x):
if i%2 == 0:
num.append(i)
else:
for i in range(x,y):
if i%2 == 1:
num.append(i)
return num
print(number(2,10))
print(number(10,2))
The outputs are:
[3, 5, 7, 9]
[2, 4, 6, 8]
Let me know if this doesn't serve your purpose.
Upvotes: 1