Question

How would you add a constant number, say 1, to a value in a dictionary if certain conditions are fulfilled.

For example, if I had a dictionary:

dict = {'0':3, '1':3, '2':4, '3':4, '4':4}

If I simply wanted to add the integer 1 to every value in the dictionary so it updates dict as this:

dict = {'0':4, '1':4, '2':5, '3':5, '4':5}

When I used the following code where the Cur_FID is the first one in the dictionary ‘0’, it gave me a value of 5? It should have given me 4. ??

for lucodes in gridList2:   # a list of the values [3,3,4,4,4] -- have to separate out because it's part of a larger nested list
    if lucodes > 1:
        if lucodes < 5:
            FID_GC_dict[Cur_FID] = lucodes + 1

print FID_GC_dict[Cur_FID]   #returned 5??? weird

I want to add 1 to all the values, but stopped here when the first dictionary update did something weird.

Answer 1

One simple way to do this is to use a collections.Counter object, which you can use in every way like a normal dictionary in most ways but it is optimized for keeping a count of items:

>>> from collections import Counter
>>> d = Counter({'0':3, '1':3, '2':4, '3':4, '4':4})
>>> d
Counter({'3': 4, '2': 4, '4': 4, '1': 3, '0': 3})
>>> d.update(d.keys())
>>> d
Counter({'3': 5, '2': 5, '4': 5, '1': 4, '0': 4})

As for only doing it when certain conditions are fulfilled, just use a comprehension or generator to only pass the list of the keys you want to increment to d.update():

>>> d = Counter({'3': 4, '2': 4, '4': 4, '1': 3, '0': 3})
>>> d.update((k for k, v in d.items() if v == 4))
>>> d
Counter({'3': 5, '2': 5, '4': 5, '1': 3, '0': 3})