Python Logic - Simplify/more efficient multiple if elif statements

Question

I'm taking an Intro to Python course online and would like to solve this problem more efficiently.

The words 1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th are called ordinal adjectives. Write a program which reads an integer x between 1 and 9 from input. The program should output the ordinal adjective corresponding to x. Hint: you don't need to have 9 separate cases; 4 is enough.

x = int(input())
if x == 1:
   print("1st")
elif x == 2:
   print("2nd")
elif x == 3:
   print("3rd")
elif x == 4:
   print("4th")
elif x == 5:
   print("5th")
elif x == 6:
   print("6th")
elif x == 7:
   print("7th")
elif x == 8:
   print("8th")
elif x == 9:
   print("9th")

I can't figure out how to write this more efficiently in four lines. Does it involve lists? They have not taught lists or arrays yet.

score 8 · Accepted Answer · answered Sep 09 '13 at 06:39

8

x = int(input())
if x == 1:
   print("1st")
elif x == 2:
   print("2nd")
elif x == 3:
   print("3rd")
else:
   print(str(x)+"th")

answered Sep 09 '13 at 06:39

freakish

54,167
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Greg · Answer 2 · 2013-09-12T06:46:52.353

Just use string formatting. And account for different suffixes between 1, 2, 3, and the rest of the numbers

x = int(raw_input())
if x == 1:
    suffix = "st"
elif x == 2:
    suffix = "nd"
elif x == 3:
    suffix = "rd"
else:
    suffix = "th"
print "{number}{suffix}".format(number=x,suffix=suffix)

You could actually do it in a more pythonic manner with a mapping instead of an if/else block

x = int(raw_input())
suffix = {1: "st", 2: "nd", 3: "rd"}.get(x, "th")
print "{number}{suffix}".format(number=x, suffix=suffix)

score 2 · Answer 3 · answered Sep 09 '13 at 06:43

2

Why convert to int?

x = input()
if x == "1":
    print("1st")
elif x == "2":
    print("2nd")
elif x == "3":
    print("3rd")
else:
    print(x+"th")

answered Sep 09 '13 at 06:43

Tim Pietzcker

328,213
58
503
561

score 1 · Answer 4 · answered Oct 16 '13 at 19:02

I needed my function to go above 9, but none of the answers were exactly what I needed. I came up with this and thought I'd share. I know it's not the most elegant solution, but it worked for what I needed it to do.

def ordinal_adjective(n):
    if ('%s' % (n,))[-2:] in ['11','12','13']:
        return '%sth' % n
    elif ('%s' % (n,))[-1] == '1':
        return '%sst' % n
    elif ('%s' % (n,))[-1] == '2':
        return '%snd' % n
    elif ('%s' % (n,))[-1] == '3':
        return '%srd' % n
    else:
        return '%sth' % n

x = input()
print ordinal_adjective(x)

score -1 · Answer 5 · edited May 23 '17 at 12:33

-1

you can't do switch case statement in python, but you can use these examples to do it in a different way, not necessary better:

here
here

edited May 23 '17 at 12:33

Community

1
1

answered Sep 09 '13 at 06:40

No Idea For Name

11,411
10
42
70

1

It might be useful to add a short summary for each. – glglgl Sep 09 '13 at 06:42
If you add a summary, your post could continue to be useful after your links go dead. :) – Jordan Sep 11 '13 at 17:45

Python Logic - Simplify/more efficient multiple if elif statements

5 Answers5