How to increment the values in a dictionary without using a loop?-CodePudding

For example I have the following set up:

dic = {"A":0, "B":0, "C":0} 
tokens = ["A", "B", "C", "C", "D", "E", "F"]

If the element in the token exists in the dictionary keys, increment the value by 1.

How do I do this without using the loop? I have the following for loop right now

for key in dict.keys(): 
    if key in tokens: 
         dict[key]  = 1

CodePudding user response：

This is a no-loop version using Counter:

from collections import Counter

dic = {"A": 0, "B": 0, "C": 0}
tokens = ["A", "B", "C", "C", "D", "E", "F"]

res = Counter(filter(dic.__contains__, tokens))
print(res)

Output

{'C': 2, 'A': 1, 'B': 1}

UPDATE

If there is a key in dic, that is not present in tokens, you could do:

from collections import Counter

dic = {"A": 0, "B": 0, "C": 0, "Z": 0}
tokens = ["A", "B", "C", "C", "D", "E", "F"]

dic.update(Counter(filter(dic.__contains__, tokens)))
print(dic)

Output

{'A': 1, 'B': 1, 'C': 2, 'Z': 0}

CodePudding user response：

You can use a recursive approach with update the dictionary by slicing one by one each tokens.

dic = {"A":0, "B":0, "C":0} 
tokens = ["A", "B", "C", "C", "D", "E", "F"] 

def key_counter(d, tokens):
    if tokens == []:
        return
    if tokens[0] in d:
        d[tokens[0]]  = 1

    key_counter(d, tokens[1:])

key_counter(dic, tokens)

print(dic)

Output

{'A': 1, 'B': 1, 'C': 2}

CodePudding user response：

Assuming that initial values in dic are always zero, you can map the dictionary keys to the tokens's count method and reform the dictionary using zip:

dic = {"A":0, "B":0, "C":0} 
tokens = ["A", "B", "C", "C", "D", "E", "F"]

dic = dict(zip(dic,map(tokens.count,dic)))
print(dic)
# {'A': 1, 'B': 1, 'C': 2}

This way, you have no import library and no for-loop (not even inside a comprehension).

but if you are concerned about performance and don't mind using a library, I would suggest using the Counter class instead of the list's .count() method:

 dic = dict(zip(dic,map(Counter(tokens).__getitem__,dic)))

If initial values in dic can be other than zero, you will need to add them in:

dic = dict(zip(dic,map(lambda k:dic[k] tokens.count(k),dic)))
# or 
dic = dict(zip(dic,map((Counter(dic) Counter(tokens)).__getitem__,dic)))

Alternatively, you can write a recursive function (but that is much slower and will be limited by the maximum recursion depth):

def countTokens(counts,token,*more):
    if token in counts: counts[token]  = 1
    if more: countTokens(counts,*more)

dic = {"A": 0, "B": 0, "C": 0}
tokens = ["A", "B", "C", "C", "D", "E", "F"]

countTokens(dic,*tokens)
print(dic)
{'A': 1, 'B': 1, 'C': 2}

CodePudding user response：

You need to iterate on tokens to read them all, not only on the keys you already set

dic = {"A": 0, "B": 0, "C": 0}
tokens = ["A", "B", "C", "C", "C", "D", "E", "F"]

for token in tokens:
    if token in dic:
        dic[token]  = 1

print(dic)  # {'A': 1, 'B': 1, 'C': 3}

Without a for loop you could use collections.Counter, but that doesn't filtering

dic = Counter(tokens)
print(dic)  # {'C': 3, 'A': 1, 'B': 1, 'D': 1, 'E': 1, 'F': 1}

CodePudding user response：

you can use comprehension

dic = { k:v tokens.count(k) for k,v in dic.items() }

but this is inefficient if the dictionary has a large number of items and tokens is a small list

CodePudding user response：

= {"A":0, "B":0, "C":0} tokens = ["A", "B", "C", "C", "D", "E", "F"] 1# convert list to set tokens 2# convert set to list 3# sort list 4# delete character D , E , F 5# convert list to tuples 6# convert to tuple to dictionary and apply dictionary 7# simple trick you take form user input