python - problem with output - iterate over dictionary

Question

I want to print('Doesn't exist') when the input is not a key in dictionary, it looks like its checking every single key in dictionary and prints if its there or not. I would like to ask if its possible to get output only one time if dictionary doesn't' contain input.

number = input('Number: ')

data = {
        '1' : 'One',
         '2' : 'Two',
         '3' : 'Three',

        }

number = number.upper()
for a, value in data.items():
    if a == number:
        print(number, ":", value)
    else:
         print('Doesnt exist here')

Number: 1
1 : One
Doesnt exist here
Doesnt exist here

You don't need a loop. `if number in data: print data[number]` — John Gordon, Sep 12 '19 at 23:33
Your code doesn't do what you describe. It scans all the items IN the dictionary, therefore they are known to exist. You are checking against a number and for all the items that DO exist you print "don't exit here". The python dictionaries have a built-in functionality to check whether a key exists. You can use it as **pppery** has suggested. If you want to do the exercise of iterating over a collection you should stop on the fist found item and either keep a flag, or break the code flow (like return from a function) — Dr Phil, Sep 12 '19 at 23:33

score 0 · Answer 1 · answered Sep 12 '19 at 23:35

0

Don't use the upper() function. Keys aren't chars.

answered Sep 12 '19 at 23:35

ttppbb

65
8

Removing the `.upper()` call would not fix the issue mentioned in the question. – pppery Sep 12 '19 at 23:38
yep! thanks, will remember, I wanted to use it with strings. In this case I have just numbers, so no need to use .upper() – Edyta Sep 12 '19 at 23:41
@pppery Symbols and Numbers are ignored by upper() python function. – ttppbb Sep 12 '19 at 23:47

score 0 · Answer 2 · answered Sep 12 '19 at 23:43

0

So you do not need to use for loop. instead of for loop add this line:

print(number, ":", data[number]) if number in data else print("Doesn't exist")

also eliminate that upper method since it is ineffective.

Have fun!

answered Sep 12 '19 at 23:43

Mohsen Abedin Nejad

81
1
3

score 0 · Accepted Answer · answered Sep 12 '19 at 23:47

There are multiple possible solutions to this problem:

use in keyword

number = input('Number: ')

data = {
        '1' : 'One',
         '2' : 'Two',
         '3' : 'Three',
}

if number in data:
    print("{}: {}".format(number, data[number]))
else:
    print('Doesnt exist here')

OR Doc

The python dictionaries throws KeyError if the key is not in

try:
    print("{}: {}".format(number, data[number]))
except KeyError as ex:
    print('Doesnt exist here')

OR Doc

print(data.get(number, "Does not exist here"))

Hope this helps, Happy learning.

score 0 · Answer 4 · answered Sep 12 '19 at 23:51

0

data = {
        '1' : 'One',
         '2' : 'Two',
         '3' : 'Three',
}
number = input('Number: ')
str_number = str(number)
if str_number in data:
    print(str_number, ":", data[str_number])
else:
     print('Doesnt exist here')

answered Sep 12 '19 at 23:51

Massifox

4,369
11
31

python - problem with output - iterate over dictionary

4 Answers4