Here's my code:

msgs = {}
msgs['ch1']=['testmsg1', 'testmsg2']
msgs['ch2']=['testmsg3','testmsg4','testmsg5']

counter = 0
for key, val in msgs.items():
    while counter < len(val):
        print(key, val[counter])
        counter  = 1

Here's the output I'm currently getting:

ch1 testmsg1
ch1 testmsg2
ch2 testmsg5

And here's the output I'd like:

ch1 testmsg1
ch1 testmsg2
ch2 testmsg3
ch2 testmsg4
ch2 testmsg5

Thanks in advance for your help, I'm still learning and would benefit from an explanation!

CodePudding user response：

It looks like you want to get all the values in the list. As @Jnevill said, you could fix how you're using the counter, but since you know how many objects are in the list you would be better to use a for loop instead of a while loop. So

msgs = {}
msgs['ch1']=['testmsg1', 'testmsg2']
msgs['ch2']=['testmsg3','testmsg4','testmsg5']

for key, val in msgs.items():
    for element in val:
        print(key, element)

CodePudding user response：

If you want to use the while loop here, you need to move your counter variable inside the for loop like such:

msgs = {}
msgs['ch1']=['testmsg1', 'testmsg2']
msgs['ch2']=['testmsg3','testmsg4','testmsg5']

for key, val in msgs.items():
    counter = 0
    while counter < len(val):
        print(key, val[counter])
        counter  = 1

This way, your counter will reset to 0 when it goes to the next key in the dictionary. Previously, the counter is 0 at testmsg1, 1 at testmsg2, but when it encounters the key ch2, the counter is now at 2, and only prints testmsg5.

CodePudding user response：

Can you do the following, or am I missing something?

msgs = {}
msgs['ch1']=['testmsg1', 'testmsg2']
msgs['ch2']=['testmsg3','testmsg4','testmsg5']

for key in msgs:
    for val in msgs[key]:
        print(key, val)

CodePudding user response：

In a single iteration:

• make a string template
• use map to apply the template to each value in the list
• use print with separator \n

for k, vs in msgs.items():
    template = k   ' {}'
    print(*map(template.format, vs), sep='\n')

Or with a simple 2-lines way using str.join:

for k, vs in msgs.items():
    print('\n'.join(f'{k} {v}' for v in vs))