Python nested loops with if statements: Replacing values inside a range

Question

Question

Given a range from 0 to 30, output all the integers inside the range with the following requirements:

I. Replace all the numbers that aren't divisible by 3 with Asterisk symbol(*);

II. Replace all the numbers inside range(10, 15) and range(20, 25) with Underscores(_).

Sample Output

0 * * 3 * * 6 * * 9 _ _ _ _ _ _ * * 18 * _ _ _ _ _ _ * 27 * * 30

My Codes:

range1 = range(10, 16)
range2 = range(20, 26)
for i in range(0, 31):
    while i not in range1 or range2:
        if i % 3 == 0:
            print(i, end = " ")
            break
        elif i % 3 != 0:
            print('*', end = " ")
            break
    else:
        print('_', end = " ")

My Output:

0 * * 3 * * 6 * * 9 * * 12 * * 15 * * 18 * * 21 * * 24 * * 27 * * 30

I am struggling with my codes since I've tried many times my codes failed to replace the numbers inside both ranges, it always skips the range check and outputs *. Any helpful ideas are appreciated, I hope I can improve my logic mindset instead of just getting the answers.

mozway · Accepted Answer · 2022-07-08T15:07:51.127

2

You can't use i not in range1 or range2, I suggest to use set union for efficiency. Also the while loop is useless, a simple test is sufficient.

NB. trying to keep an answer as close as possible to your original code here.

ranges = set(range(10, 16)) | set(range(20, 26))

for i in range(0, 31):
    if i not in ranges:
        if i % 3 == 0:
            print(i, end = " ")
        elif i % 3 != 0:
            print('*', end = " ")
    else:
        print('_', end = " ")

output: 0 * * 3 * * 6 * * 9 _ _ _ _ _ _ * * 18 * _ _ _ _ _ _ * 27 * * 30

ranges: {10, 11, 12, 13, 14, 15, 20, 21, 22, 23, 24, 25}

edited Jul 08 '22 at 15:07

answered Jul 08 '22 at 15:01

mozway

194,879
13
39
75

Thank you for the quick comment, any methods besides the ```set``` function? Because it's beyond my current studying content. – Jul 08 '22 at 15:05
1

You can use `ranges = list(range(10, 16)) + list(range(20, 26))` instead, this gives you a list, less efficient but it will work ;) – mozway Jul 08 '22 at 15:06

score 1 · Answer 2 · answered Jul 08 '22 at 15:06

1

It's so easy and simple to do it without extra ranges:

for i in range(0, 31):
    if (10 <= i <= 16) or (20 <= i <= 26):
        print('_', end=' ')
    elif i % 3 != 0:
        print('*', end=' ')
    else:
        print(i, end=' ')

answered Jul 08 '22 at 15:06

QwertYou

333
2
12

Thank you, this is very close to my current knowledge. – Jul 08 '22 at 15:12
1

You are welcome! Good luck in your studies! – QwertYou Jul 08 '22 at 15:13

Timur Shtatland · Answer 3 · 2022-07-08T16:34:04.157

Do not use range, use a simpler and more efficient syntax of chained comparisons instead. Simplify your if condition as well: there is no need to check twice whether the number is divisible by 3. Skip break too. Separate creation of the list using append and printing of the list, which you do after the list is created (creation of the data and its printing are best kept separate).

lst = []
for i in range(31):
    if 10 <= i <= 15 or 20 <= i <= 25:
        lst.append('_')
    elif i % 3:
        lst.append('*')
    else:
        lst.append(i)

print(' '.join(str(x) for x in lst))
# 0 * * 3 * * 6 * * 9 _ _ _ _ _ _ * * 18 * _ _ _ _ _ _ * 27 * * 30

# You can also use a simpler `print` to give the same result, 
# as suggested by *QwertYou*:
print(*lst)

`print(*lst)` looks less bulky – QwertYou Jul 08 '22 at 15:12 — QwertYou, Jul 08 '22 at 15:12

Python nested loops with if statements: Replacing values inside a range

3 Answers3