places = ["Jack", "John", "Sochi"]
count=0
multi_word=0
place = places[count]
while place != "Sochi" and count < len(places):
if ' ' in place:
multi_word += 1
count += 1
place = places[count]
print ('Number of cities before Sochi:', count)
My code should print the number of cities before Sochi excluding Sochi . I don't understand what this line (place = places[count]) does, nor do I understand why I need it twice.
foreach would neaten it up
places = ["Jack", "John", "Sochi"]
count = 0
for place in places:
if ' ' in place:
multi_word += 1
if place == "Sochi":
break
count += 1
count=0
place = places[count]
Now place is always places[0], i.e. Jack. Thus the while loop only terminates on the second condition, giving you the list length of 3.
place = places[count] should go in the loop.
You can use the following while loop to check for the number of places before Sochi:
places = ["Jack", "John", "Sochi"]
count = 0
multi_word = 0
while count < len(places):
place = places[count]
if ' ' in place:
multi_word += 1
if place == "Sochi":
break
count += 1
print('Number of cities before Sochi:', count)
The break statement means you'll exit your while loop.
count < len(places) condition. If "Sochi" is not in the list, you'll get out of bounds after the end.Why not try a more pythonic solution instead ?
places = ["Jack", "John", "Sochi"]
try:
count = places.index("Sochi")
except ValueError:
count = len(places)
multi_word = len([place for place in places[:count] if ' ' in place])
**place = places[count]**?placebefore used in the while loop.places.index('Sochi')which will return the index for the string 'Sochi' in the list and since lists are zero-indexed you'll get the sought after number