Python update list of dictionaries in a loop

Question

I am working on a python code to update list of dictionaries if input exists in list of dictionaries. If input doesn't exist in list of dictionaries it should print "value doesn't exist in entire list" or do some other operation. Below is the code I have written

a = [{'main_color': 'red', 'second_color':'blue'},{'main_color': 'yellow', 'second_color':'green'},{'main_color': 'yellow', 'second_color':'blue'}]

print('Enter main color:')

conType=input()

for d in a:

    if d['main_color']==conType:
        print('matched')
        d['second_color']='bluetooth'
    else:
        print('no value')
print(a)

Issue here is if input is 'red' "no value" is getting printed twice and "matched" is printed once.

My use case is if input doesn't exist in list of dictionaries it should print "no value" only once. If input exists code should update next key with new value and print "matched" once.

I have been through other questions in stack overflow. I couldn't find answer to this scenario.

Please help

Tried. Now it's printing matched once, no value once – sudhir May 28 '19 at 16:07 — sudhir, May 28 '19 at 16:07

Olvin Roght · Accepted Answer · 2019-05-29T10:03:16.980

0

There's for .. else statement which you can use here:

a = [{'main_color': 'red', 'second_color': 'blue'},
     {'main_color': 'yellow', 'second_color': 'green'},
     {'main_color': 'yellow', 'second_color': 'blue'}]

conType = input('Enter main color: ')

for d in a:
    if d['main_color'] == conType:
        print('matched')
        d['second_color'] = 'bluetooth'
        break
else:
    print('no value')

print(a)

UPD.

Sorry, I've missed that there's multiple objects with same main_color in the list (thanks, @HaleemurAli). In case if you don't need to break loop after matching you can't use for .. else statement, but you can use boolean flag. To check if string equals any of list of string you can membership test operator (in):

a = [{'main_color': 'red', 'second_color': 'blue'},
     {'main_color': 'yellow', 'second_color': 'green'},
     {'main_color': 'blue', 'second_color': 'blue1'}]

conType = input('Enter main color: ')
conType1 = input('Enter another main color: ')

b = [conType, conType1]
matched = False
for d in a:
    if d['main_color'] in b:
        print('matched')
        matched = True
        d['second_color'] = 'bluetooth'

if not matched:
    print('no value')

edited May 29 '19 at 10:03

answered May 28 '19 at 16:08

Olvin Roght

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Thanks. This is Interesting. Learned new concept. I am not aware that else statement can be written outside loop – sudhir May 28 '19 at 16:10
this won't update the 3rd dictionary if `conType == 'yellow'` – Haleemur Ali May 28 '19 at 16:20
@Olvin Roght What if there are multiple inputs. How do we loop in that case? a = [{'main_color': 'red', 'second_color': 'blue'}, {'main_color': 'yellow', 'second_color': 'green'}, {'main_color': 'blue', 'second_color': 'blue1'}] conType = input('Enter main color: ') conType1= input('Enter another main color: ') b=[] b.append(conType) b.append(conType1) for d in a: for i in b: if d['main_color'] == i: print('matched') d['second_color'] = 'bluetooth' break else: print('no value') – sudhir May 29 '19 at 07:20
@HaleemurAli, thanks, I've missed that. Edited code. – Olvin Roght May 29 '19 at 10:03
@sudhir, check post edits. – Olvin Roght May 29 '19 at 10:04

Python update list of dictionaries in a loop

1 Answers1