How to create an if statement for elements within a list?

Question

I am wondering how I can create an if statement that does something to a list as long as there is at least one value in the list that meets the if statement's requirements.

For example, given a list

x=[1,2,3,4,5,6,7,8,9]
y=[1,2,3,4]
z=[1,2,3,8]

I want to create an if statement where if the list contains elements whose values are between 5 and 10, then the list appends ***.

Meaning, after the if statement, the results would be

x=[1,2,3,4,5,6,7,8,9,***]
y=[1,2,3,4]
z=[1,2,3,8,***]

since both x and z contains elements that are between 5 and 10, they receive the ***.

• Sheldore: What is ***??
• John Coleman: Perhaps this will help: stackoverflow.com/q/19389490/4996248
• IanQ: This sounds like homework. At the very least you should show us what you have tried
• Valentino: You should loop over the list (or use a list comprehension) to verify that the condition is satisfied and in case do what you want to do.

Answer 1

You could do this. Test if any of the elements match your condition using a generator expression and the any() function.

x = [1,2,3,4] #two lists for testing
y = [5] 
if any(5 <= i <= 10 for i in x):
 x.append("***")
if any(5 <= i <= 10 for i in y):
 y.append("***")
print(x,y)

Output:

[1, 2, 3, 4] [5, '***']

Answer 2

The most succinct way to do this in Python 3 is to use a generator expression:

if any(5 < x < 10 for x in lst):
 lst.append('***')

Here is a working example.

Edit: That syntax is kind of mind blowing, thanks for the edit and the comment.

Answer 3

i=0
while i<len(x):
 if x[i]>5 and x[i]<10: #if any item of list is between 5 and 10 break loop
 break
 i+=1
if i<len(x): #check if i less than len(x) this means that loop 'broken' until the end
 x.append('***')

• Dorian Turba: You should try to do this without while, iterator and break. Your code is too complex. Do not forget the Zen of python and Keep It Simple Stupid (KISS).