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How dictionaries behave

Here's the code:

checkDictionary =
'0': (print('0'),0),
'1': (print('1'),1),
'2': (print('2'),2)



print(checkDictionary.get('1','Not found'))


The output I expected was:

1
(None, 1)


but instead it runs all the functions in every value field:

0
1
2
(None, 1)


Can anyone explain this behavior, and why was it implemented like this?

I heard from multiple sources that Python dictionaries are supposed to be O(1) fast "most of the time".

So the thing I really wanted to accomplish was to make a nice switch/case with multiple executing lines:

def readPoint(v):
print(f'Point: v')

def readLine(v):
print(f'Line: v')

Point = 1
Line = 2
segment = [1,1]
checkDictionary =
Point: "geometry = readPoint(segment)n__type='Point'",
Line: "geometry = readLine(segment)n__type='Line'"



exec(checkDictionary.get(segment[0],'raise ValueError("Some error")'))


Would there be some issues with this design pattern? don't care about the functions or variables they are only examples.

checkDictionary

'0': (None, 0), '1': (None, 1), '2': (None, 2), '3': (None, 3), '4': (None, 4), '5': (None, 5), '6': (None, 6), '7': (None, 7)

[(print('0'), 0)]

(None, 0)

2 Answers
2

This is totally expected and in no way a strange behavior. The key to understanding is knowing that the dictionary gets created (keys and values are evaluated) when you declare it.

In your case, the evaluation of the values results in the execution of the print statements.

print

This is why you get:

0
1
2
3
4
5
6
7


Then comes print(checkDictionary.get('1','Not found'))

print(checkDictionary.get('1','Not found'))

which expectedly produces:

(None, 1)


lambda

functools.partial

It does it because anywhere you're calling print it will print, so the best way to solve this is to remove the print or something else and also print is a none type

print

print

Actually (None, 1) is the output of the code, it's just that when you're creating the dictionary it is printing

(None, 1)

so at the end your code is working as expected

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