How do i make python find differing elements in a list and their number of occurrences within it-CodePudding

Say i have a list x and : x = ['a', 'b', 'c', 'c', 'c', 'c', 'a']

if a user, or another programmer, does not know the elements in a list, how do I tell python to find the unique elements So if they ask (what is in this list?) python would output: in this list there are 3 different elements : a, b and c

and if the user asks (how many of each) python should output: ie - (in this list, there are 4 instances of c, 2 instances of a and 1 instance of b)

CodePudding user response：

Look at collections.Counter.

x = ['a', 'b', 'c', 'c', 'c', 'c', 'a']
counter = collections.Counter(x)
print(len(counter))  # 3
print(counter)  # Counter({'c': 4, 'a': 2, 'b': 1})

CodePudding user response：

I propose:

ItemList = ['a', 'b', 'c', 'C', 'c', 'C', 'A']
NewDict = {}

for Item in ItemList:
  if Item not in NewDict:
    NewDict[Item] = 0
  NewDict[Item]  = 1

print(NewDict)

And if you don't want respect the case:

ItemList = ['a', 'b', 'c', 'C', 'c', 'C', 'A']
NewDict = {}

for Item in ItemList:
  if Item.lower() not in NewDict:
    NewDict[Item.lower()] = 0
  NewDict[Item.lower()]  = 1

print(NewDict)

CodePudding user response：

Here's a quick solution without collections.Counter:

x = ['a', 'b', 'c', 'c', 'c', 'c', 'a']
print(len(set(x)))
print([f"{i} - {x.count(i)}" for i in set(x)])