在字典中查找特定值，Python

Question

position={'Part1':('A23-1','A24-2','A24-4','A25-1','A27-5'),
          'Part2':('A26-7','B50-6','C1-3'),
          'Part3':('EM45-4','GU8-9','EM40-3','A15-2')}

So I have this dictionary, showing a "part" as the key and the value being the position in a warehouse. Now let's say I want to find which parts are on shelf A25 through A27, I've met a wall. So far I've come up with:

for part, pos in position:
    if str.split(pos)=='A25' or 'A26' or 'A27':
        print(part,'can be found on shelf A25-A27')

However, this gives me a ValueError, which I've found to be 'cause all of the values have different lengths, so how do I go about this?

Answer 1

I'm not sure what you think str.split(pos) is going to do, but probably not what you want. :)

Similarly, if foo == 1 or 2 doesn't check whether foo is either of those values; it's parsed as (foo == 1) or 2 , which is always true because 2 is a true value. You want foo in (1, 2) .

You're also trying to loop over position alone, which gives you only the keys; that's probably the source of your error.

The values in your dictionary are themselves tuples, so you need a second loop:

for part, positions in position.items():
    for pos in positions:
        if pos.split('-')[0] in ('A25', 'A26', 'A27'):
            print(part, "can be found on shelves A25 through A27")
            break

You could avoid the inner loop with an any as shown in the other answer, but imo that's hard to read with a complex condition like this one.

Answer 2

This is what you're trying to do:

>>> position={'Part1':('A23-1','A24-2','A24-4','A25-1','A27-5'),
...           'Part2':('A26-7','B50-6','C1-3'),
...           'Part3':('EM45-4','GU8-9','EM40-3','A15-2')}
>>> for part,pos in position.items():
...     if any(p.split('-',1)[0] in ["A%d" %i for i in range(25,28)] for p in pos):
...             print(part,'can be found on shelf A25-A27')
... 
Part1 can be found on shelf A25-A27
Part2 can be found on shelf A25-A27

But I'm going to suggest that you transform your data structure a little:

>>> positions = {}
>>> for item,poss in position.items():
...     for pos in poss:
...         shelf, col = pos.split('-')
...         if shelf not in positions:
...             positions[shelf] = {}
...         if col not in positions[shelf]:
...             positions[shelf][col] = []
...         positions[shelf][col].append(item)
... 

Now you can do this:

>>> shelves = "A25 A26 A27".split()
>>> for shelf in shelves:
...     for col in positions[shelf]:
...         for item in positions[shelf][col]:
...             print(item, "is on shelf", shelf)
... 
Part1 is on shelf A25
Part2 is on shelf A26
Part1 is on shelf A27

Answer 3

Here is an efficient one-line solution:

>>> position = {'Part1':('A23-1','A24-2','A24-4','A25-1','A27-5'),
...             'Part2':('A26-7','B50-6','C1-3'),
...             'Part3':('EM45-4','GU8-9','EM40-3','A15-2')}
>>> lst = [k for k,v in position.items() if any(x.split('-')[0] in ('A25', 'A26', 'A27') for x in v)]
>>> lst
['Part2', 'Part1']
>>>
>>> for x in sorted(lst):
...     print(x,'can be found on shelf A25-A27.')
...
Part1 can be found on shelf A25-A27.
Part2 can be found on shelf A25-A27.
>>>