ex3.py

txt="Les objectifs de Python en font un langage pedagogique ideal.C'est un langage     general-purpose:il s'adapte a toutes les applications"

seperatorList = " ,.;-'\":\n/\\?+()’!°"

n_words = 0
words   = []

while len(txt) > 0:
    seperation = len(txt)
    for sep in seperatorList:
        sepPos = txt.find(sep)
        if sepPos >= 0 and sepPos < seperation:
            seperation = sepPos
    if seperation > 0:
        n_words+=1
        txt=txt[seperation+1:]
    elif seperation == 0:
        txt=txt[seperation+1:]
    else:
        words.append(txt)
        txt=""

print("Il y a {0} mots dans la phrase.".format(n_words))

# solution alternative sans utiliser "find"
# moins optimale sur de grands textes

in_word = False
words = []
curword = ""

for letter in txt:
    if letter in separators:
        if in_word:
            words.append(curword)
            curword = ""
            in_word = False
    else:
        in_word = True
        curword += letter

if in_word:
    words += curword

print("il y a", len(words), "mot(s)")