Programming Praxis

6 Responses to “First Word”

1. matthew said

   January 25, 2019 at 10:46 AM

   Extra points for implementing the Unicode default word breaking algorithm: https://unicode.org/reports/tr29/#Word_Boundaries
2. V said

   January 25, 2019 at 12:14 PM

   Two solutions in golang.

   
   package main
   
   import (
   	"fmt"
   	"regexp"
   )
   
   func main() {
   	s1 := "abcdefg hijklmnop qrs tuv wxyz"
   	s2 := "~hola   caracola  de piazzolla"
   	fmt.Println(firstWordWithLoop(s1))
   	fmt.Println(firstWordWithRegexp(s1))
   	fmt.Println()
   	fmt.Println(firstWordWithLoop(s2))
   	fmt.Println(firstWordWithRegexp(s2))
   }
   
   func firstWordWithLoop(str string) string {
   	word := ""
   	for _, char := range str {
   		if char == ' ' {
   			break
   		}
   		word += string(char)
   	}
   	return word
   }
   
   func firstWordWithRegexp(str string) string {
   	return regexp.MustCompile(`[^ ]+`).FindString(str)
   }
   
   
3. Daniel said

   January 25, 2019 at 4:46 PM

   Here’s a solution in C.

   #include <stdio.h>
   #include <stdlib.h>
   
   int main(int argc, char* argv[]) {
     if (argc != 2) {
       fprintf(stderr, "Usage: %s STR\n", argv[0]);
       return EXIT_FAILURE;
     }
     char* str = argv[1];
     while (1) {
       char c = *(str++);
       if (c == ' ' || c == '\0') break;
       printf("%c", c);
     }
     printf("\n");
     return EXIT_SUCCESS;
   }
   

   Example Usage:

   $ ./a.out "abcdefg hijklmnop qrs tuv wxyz"
   abcdefg
   
4. Steve said

   January 25, 2019 at 6:08 PM

   AWK version

   $ echo "abcdefg hijklmnop qrs tuv wxyz" | awk ‘{ print $1 }’
   abcdefg

   —

   Klong version

   (-1)_((a?" ")@0)#a::"abcdefg hijklmnop qrs tuv wxyz"
   "abcdef"
   a
   "abcdefg hijklmnop qrs tuv wxyz"
   a?" "
   [7 17 21 25]

   —

   MUMPS version

   YDB>w $p("abcdefg hijklmnop qrs tuv wxyz"," ")
   abcdefg
5. matthew said

   January 25, 2019 at 11:41 PM

   That Unicode algorithm looks a bit complicated, here’s a simple Unicode-friendly solution using Python str.isspace():

   def firstword(s):
       start = -1
       for i,c in enumerate(s):
           if start < 0:
               if not c.isspace(): start = i;
           elif c.isspace():
               return s[start:i]
       return None if start < 0 else s[start:]
   
   assert(firstword("") is None)
   assert(firstword("  ") is None)
   assert(firstword("foo") == "foo")
   assert(firstword(" foo") == "foo")
   assert(firstword("foo ") == "foo")
   
6. Andrey Sidorenko said

   February 8, 2019 at 1:32 PM

   Python solution:

   firstWord = lambda x: x.lstrip(" ").split(" ")[0]