创建一个列表，计算另一个列表中数字的连续出现次数

Question

I have this list

x = [2,3,4,2,2]

I want to create a new list of the same size where each element will represent a count of the number of times the corresponding number has appeared in the original list before the current element. For example, the first 3 elements are 2,3,4 So, I want the first 3 values in the new list to be 0, and in x[3] since 2 is repeated, I want the value to be 1 and x[4] it's again 2 I want the value to be 2 (the number 2 has been encountered in the list twice up to this point).

My expected answer is the new list = [0,0,0,1,2]

Similarly for the list [2,3,4,2,2,3,3] I want the new list to be [0,0,0,1,2,1,2]

Answer 1

You can leverage itertools.count and collections.defaultdict to make a simple, efficient list comprehension. By setting count() as the default value for your default dictionary you get a new counter item for every unique number in the list that increments as you call next() on it.

from itertools import count
from collections import defaultdict

c = defaultdict(count)
x = [2,3,4,2,2]

[next(c[n]) for n in x ]
# [0, 0, 0, 1, 2]

Answer 2

You can keep track of the number of times a number has been met and add it to the result accordingly.

>>> d = defaultdict(int)
>>> res = []
>>> for el in x:
...     res.append(d[el])
...     d[el]+=1
... 
>>> res
[0, 0, 0, 1, 2]

Answer 3

I don't see any shortcuts to solving this other than full exploration.

counters = {} # keep track of counts
x= [2,3,4,2,2]
new_list = [counters[c] for c in x if not counters.update({c:counters.get(c,-1)+1})]

Answer 4

I came up with this one:

print([x[:i].count(x[i]) for i in range(len(x))])

First iterate over the index of the list: for i in range(len) Then take the value at position i of the original list and count how many times it occurs before that position. For the original list x = [2,3,4,2,2] we obtain:

i=0  [].count(2) -> 0
i=1  [2].count(3) -> 0
i=2  [2,3].count(4) -> 0
i=3  [2,3,4].count(2) -> 1
i=4  [2,3,4,2].count(2) -> 2

Answer 5

A shorter but probably less efficient solution:

[x[:i].count(n) for i,n in enumerate(x)]

Answer 6

You could try this as well:

list = [2, 3, 4, 2, 2]
repeats = []
repeatsCounter = 0
for i in range(len(list)): #loops through the list
   for j in range(i - 1, -1, -1): #loops through every value until i
      if list[i] == list[j]:
         repeatsCounter += 1
         break
   repeats.append(repeatsCounter)