CODEWITHSUNDEEP
python
Meowmeow
Meowmeow

Reputation: 23

Python - I cant get names past the first one from a list

I'm new to python and i got a problem with my code.. so what I'm trying to do is to ask some random name and that program should tell you if a name is in the list

names = ["Demi", "Siru", "Datte", "Sony", "Jami", "Windy", "Paavo", "Roosa"]
x = input('Give a name: ')
for y in names:
    if(x==y):
        print( x + " is in a list" )
        break
    elif(y!=x):
        print("it's not in a list")
        break

Upvotes: 1

Views: 43

Answers (5)

e.doroskevic
e.doroskevic

Reputation: 2169

names = ['ed', 'edd', 'eddy']
  found = False
  end = False

  while(end != True):
      user_input = input('enter name... ')

      for name in names:
          if name == user_input:
              found = True

      if found:
          print('yep! the name {} is available'.format(name))
      else: 
          print('hold up! wait a minute! no name found!')        

  again = input('continue? [y/n] ? ')

  if again == 'n':
     end = True
  else:
     found = False

Upvotes: 0

wizzwizz4
wizzwizz4

Reputation: 6426

You're iterating through the list, making a comparison, then breaking out of the loop as soon as you've processed the first item. Remove break.

That aside, the solution to your problem is to use if item in my_list: instead of that for loop nonsense.

Alternatively, you could use for: else: like so:

for y in names:
    if x == y:
        print("yes it is in there")
        break
else:
    print("no it is not")

You'll see that else: is attached to the for loop itself, as opposed to the if. The else is checking to see whether there was no break. Break causes the loop to exit early.

Upvotes: 0

zvone
zvone

Reputation: 19342

The correct code would go through all the items in the list, and if one matches break out of it, but only if none of them matched, conclude that it is not found (so that part goes out of the loop), like this:

for y in names:
    if x == y:
        print("{} is in the list".format(x))
        break
else:
    print("it's not in the list")

Of course, it can be done without the loop completely:

if x in names:
    print("{} is in the list".format(x))
else:
    print("it's not in the list")

Upvotes: 3

Haskar P
Haskar P

Reputation: 420

Use List Comprehension.

x +"In List" if x in names else "Not in List"

Upvotes: -1

Messa
Messa

Reputation: 25181

Because you check only the first item of the list and then break.

Try to remove the break.

Upvotes: 0

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