字典和嵌套字典 python

Question

Attempting to understand dicts in a somewhat simple way I believe, but am thoroughly confused. When I run the code below, it works fine to populate dict and count the occurrences of a word. Output is as such: {'gerry': 2, 'babona': 1, 'cheese': 1, 'cherry': 1}

dict = {}

a = ['gerry', 'babona', 'cheese', 'gerry', 'cherry']
b = ['O' ,'O', 'T', 'T', 'T']

for (i,j) in zip(a,b):

  if i not in dict:
    dict[i] = 1
  else:
    dict[i] += 1

However, if I try to run the following code, there is a KeyError: 'gerry' , beginning with the first value in the list, but I cannot make sense of why. Any help on this greatly appreciated!

dict = {}

a = ['gerry', 'babona', 'cheese', 'gerry', 'cherry']
b = ['O' ,'O', 'T', 'T', 'T']

for (i,j) in zip(a,b):

  if i not in dict:
    dict[i][j] = 1
  else:
    dict[i][j] += 1

Answer 1

The first example can be simpler

from collections import Counter

a = ['gerry', 'babona', 'cheese', 'gerry', 'cherry']

data = Counter(a)

For second

from collections import defaultdict


a = ['gerry', 'babona', 'cheese', 'gerry', 'cherry']
b = ['O', 'O', 'T', 'T', 'T']

data = defaultdict(lambda: defaultdict(int))

for (i, j) in zip(a, b):
    data[i][j] += 1

btw don't use reserved name for your variable eg dict

Answer 2

When the program gets to: dict[i][j] = 1 it has to execute: dict[i] first, which is exactly what you were trying to avoid when you wrote the first snippet.

You will have to do multi-stage tests to get this to work.

You can get it to work like this, but there are simpler ways:

dct = {}

a = ['gerry', 'babona', 'cheese', 'gerry', 'cherry']
b = ['O' ,'O', 'T', 'T', 'T']

for (i,j) in zip(a,b):
  if i not in dct:
    dct[i] = {}
    dct[i][j] = 1
  else:
    di = dct[i]
    if j not in di:
        di[j] = 0
    dct[i][j] += 1
print(dct)

Output:

{'gerry': {'O': 1, 'T': 1}, 'babona': {'O': 1}, 'cheese': {'T': 1}, 'cherry': {'T': 1}}

Answer 3

from collections import Counter
a = ['gerry', 'babona', 'cheese', 'gerry', 'cherry']
b = ['O' ,'O', 'T', 'T', 'T']
print(Counter(a))
Counter({'gerry': 2, 'babona': 1, 'cheese': 1, 'cherry': 1})

for the 2nd case you can use groupby

dct = dict((key, tuple(v for (k, v) in pairs)) 
           for (key, pairs) in groupby(sorted(zip(a,b)), lambda pair: pair[0]))
{k:Counter(v) for k,v in dct.items()}
{'babona': Counter({'O': 1}), 'cheese': Counter({'T': 1}),'cherry': Counter({'T': 1}), 'gerry': Counter({'O': 1, 'T': 1})}