Unexpected Behaviour of if statement inside loop over d.items() - if-statement

Question

Using Python 3.3 I want to loop over d.items() and return matches based on if condition. Here is the code:
d = {'mohammed': '123456789', 'john': '1230012'}
for k, v in d.items():
if k == 'mohammed' and v == '123456789':
print("Match")
else:
print("No Match")
I expect to print "Match", however what I get is that both the statements print in reverse, i.e.,
No Match
Match
How to get the code print only the right statement? Any help is appreciated.

Answer 1

Dictionaries are not ordered, so the entries in d may not be stored in the same order you typed them. Therefore, the order in which d.items() iterates over the entries in d cannot be predicted exactly.
When you iterate over d.items(), you are iterating over all key,value pairs. Therefore, if a pair matches your if-condition, Match will be printed. By the same reasoning, if a pair does not match your if-condition, No Match will be printed.
In your case, the order of key,value pairs during the iteration over d.items() is exactly the opposite of the order in which you've entered them. Let's take a look at this:
The first key,value pair to be considered is 'john': '1230012'. This does not match your if-condition. Therefore No Match is printed.
The next pair is 'mohammed': '123456789', which does match your if-condition. Therefore Match is printed.
You can confirm this by adding a print(k,v) before your if-statement
Now, if for some reason, you want features of dictionary, and you want to maintain the order in which you've entered your data, then I'd recommend using collections.OrderedDict
In your post, you say "I expect to print Match". I think therefore, what you're looking for, is a way to check if there are any key,value pairs in the dictionary that satisfy your if-condition. Here are two ways in which you can accomplish this:
d = {'mohammed': '123456789', 'john': '1230012'}
found = False
for k,v in d.items():
if k == 'mohammed' and v == '123456789':
print("Match")
found = True
if not found: # equivalently, if found == False:
print("No Match")
This will print Match as many times as there are key,value pairs that do match your if-condition. If however, you want to print Match only once even if multiple pairs match your if-condition, then you make a small modification to the above code:
d = {'mohammed': '123456789', 'john': '1230012'}
found = False
for k,v in d.items():
if k == 'mohammed' and v == '123456789':
found = True
if found:
print("Match")
else:
print("No Match")
Of course, there's a handy one-liner for all of this:
if any(k == 'mohammed' and v == '123456789' for k,v in d.items()):
print("Match")
else:
print("No Match")

Answer 2

Firstly, dictionaries(Dict objects) are un ordered so you can't be sure which item is accessed/iterated first.
Secondly, by the looks of it(assuming) that you want just a Match or No Match for an output.
You can do this by simple having a boolean;
matched = False
for k, v in d.items():
if k == 'mohammed' and v == '123456789':
print("Match")
matched = True
break
if not matched:
print("No Match")
OR
If you want to access these in order, you can change your data structure to something like
[('muhammad','somevalue'),(),...]
Then you may iterate on this list and have something like
for obj in d:
t_val_1, t_val_2 = obj
if t_val_1 == 'mohammed' and t_val_2 == '123456789':
print("Match")
else:
print("No Match")