通過遞增該值來修改dictionary（或defaultdictionary）中的現有值

Question

我想每次在我掃描的列表中出現特定值時都要保留一個計數器。

例如： 列表：喜歡

[(a, 0.2), (b, 1), (a, 0.2), (a, 1)] 

我的字典，可以顯示以下內容：

mydict = {"a": (# val below 1, # val equal to 1), ...}

因此： mydict = {"a": (2, 1), "b" :(0, 1)}

是有辦法用默認字典或普通字典來做到這一點？

對於每個我看到低於或等於1的值，我是否應該這樣做： mydict[mydict["a"]+1]？

Answer 1

好的，假設輸入類型是一個數組數組，並且可以將結果作爲數組存儲在字典中，那麼可以這樣做。

# Define list of numbers 
lettersNumbersList = [["a", 0.2], ["b", 1], ["a", 0.2], ["a", 1]] 

# Here is the dictionary you will populate. 
numberOccurences = {} 

# This function is used to increment the numbers depending on if they are less 
# than or greater than one. 
def incrementNumber(letter, number): 

    countingArray = numberOccurences[letter] 

    if number < 1: 
     countingArray[0] = countingArray[0] + 1 
    elif number >= 1: 
     countingArray[1] = countingArray[1] + 1 

    return(countingArray) 

# Loops through all of the list, gets the number and letter from it. If the letter 
# is already in the dictionary then increments the counters. Otherwise starts 
# both from zero. 
for item in lettersNumbersList: 

    letter = item[0] 
    number = item[1] 

    if letter in numberOccurences: 
     numberOccurences[letter] = incrementNumber(letter, number) 

    else: 
     numberOccurences[letter] = [0, 0] 
     numberOccurences[letter] = incrementNumber(letter, number) 

print(numberOccurences) 

Answer 2

這應該是比其他解決方案（也很乾淨，Python的恕我直言）快：

mylist = [("a", 0.2), ("a", 0.9), ("b", 1), ("a", 1)] 

mydict = dict(mylist) 

for k in mydict.keys(): 
    mydict[k] = (len([t for t in mylist if t[0]==k and t[1]<1]), 
       len([t for t in mylist if t[0]==k and t[1]==1])) 

# >>> mydict 
# {'a': (2, 1), 'b': (0, 1)}