Sorting strings characters in Python and test the first letter in string

Question

def a_letter(string):
  s = string.lower()
  if s[0] == 'a':
    return 'alpha'
  else:
    return 'zebra'

This code is testing the first letter of a string - if it's an 'a' then return alpha, else return zebra. Is there a better way to do this?

>is there a better way to do this What does 'better' mean to you? A solution that is faster, has less code, is more readable or what? — gonczor, Dec 28 '17 at 14:31
Here's a one liner, though I think your solution is fine: `return 'alpha' if s and s[0].lower == 'a' else 'zebra'.` — kabanus, Dec 28 '17 at 14:32
If the code is working, a better place to ask would be [codereview.se]. (Why should we have all the fun.) — Jongware, Dec 28 '17 at 14:35
still an intermediate python coder, so i think pros can have a better way than this...thanks y'all — sammiie, Dec 28 '17 at 14:37
@sammiie if you want a code review there is special stack exchange community for this (named Code Review obviously). — gonczor, Dec 28 '17 at 14:53

score 3 · Answer 1 · edited Dec 28 '17 at 14:42

3

I think a good way to do so is:

def a_letter(s):
    return 'alpha' if s and s[0].lower() == 'a' else 'zebra'

Because it is clear and sufficiently concise.

edited Dec 28 '17 at 14:42

timgeb

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answered Dec 28 '17 at 14:37

vinzee

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using slice avoids index error, so you can not check if string is empty: `return 'alpha' if string[0:1].lower() == 'a' else 'zebra'` – splash58 Dec 28 '17 at 14:48
@splash58 better yet `string[:1]` – timgeb Dec 28 '17 at 15:28

Aaditya Ura · Answer 2 · 2017-12-28T14:48:19.763

1

If you have only two conditions you can try something a poor-man's conditional expression

print(['zebra','alpha'][string[0].lower() =='a'])

How it works ?

True == 1, False == 0

because booleans are a subclass of int , so [string.lower()[0]=='a'] produce a integer value but ['false','true'] takes it as index value.

edited Dec 28 '17 at 14:48

answered Dec 28 '17 at 14:35

Aaditya Ura

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It's cute, but in the real world I'd much prefer OP's code. – timgeb Dec 28 '17 at 14:42

user2390182 · Answer 3 · 2017-12-28T14:45:12.037

There is nothing inherently wrong with being explicit. Maybe you should protect against the empty string as pointed out by Willem Van Onzem in the comments. You can do that - for instance - by using str.startswith which is the built-in way to check for prefixes. And are there shorter ways? - Sure:

return ('zebra', 'alpha')[s.startswith(('a', 'A'))]

Or maybe more readable, using the ternary operator:

return 'alpha' if s.startswith(('a', 'A')) else 'zebra'

score 0 · Answer 4 · answered Dec 28 '17 at 14:49

Do a more comprehensive solution, return something for every letter

voc= converts the list of nato alphabet names to a dict mapping of letter to name

def phonetic checks the input is in the dict and then returns the value, or ? if it is confused

voc=dict([[string.lower(x[0]), x] for x in 
['ALFA',
 'BRAVO',
 'CHARLIE',
 'DELTA',
 'ECHO',
 'FOXTROT',
 'GOLF',
 'HOTEL',
 'INDIA',
 'JULIETT',
 'KILO',
 'LIMA',
 'MIKE',
 'NOVEMBER',
 'OSCAR',
 'PAPA',
 'QUEBEC',
 'ROMEO',
 'SIERRA',
 'TANGO',
 'UNIFORM',
 'VICTOR',
 'WHISKEY',
 'XRAY',
 'YANKEE',
 'ZULU']])

def phonetic(letter):
    if letter in voc:
        return voc[letter]
    else:
        return "?"

Sorting strings characters in Python and test the first letter in string

4 Answers4