如何有效计算列表中元素的非连续出现次数？

Question

So I'm trying to find out how many times a certain element is non consecutively appearing in a list. By that I mean:

list = [10,10,10,11,12,10,12,14,10,10,10]
element_searched = 10

=> expected_output = 3

So that means that 10 is appearing 3 times in a the list.

My code so far that seems to be working:

elements = [11, 10, 12]
row = [10,10,10,10,10,10,10,10,10,11,11,11,11,11,10,10,10,10,12,12,12,12,12,11,11,11,11,12,12,12,12,10]

element_on = False
for element in elements:
    sequence = 0
    for i in range(len(row)):
        if element == row[i] and element_on==False:
            sequence += 1
            element_on = True
        elif element==row[i] and element_on==True:
            pass
        elif element != row[i] and element_on==True:
            element_on = False
        #
        elif element != row[i] and element_on == False:
            element_on = False
        #
    print(f"For element {element} the number ob sequences is: {sequence} ")

I am getting the desired output but I am wondering if there is a more elegant and especially a faster way.

Answer 1

Try this:

row = [10,10,10,10,10,10,10,10,10,11,11,11,11,11,10,10,10,10,12,12,12,12,12,11,11,11,11,12,12,12,12,10]
sr = pd.Series(row, name = "x")
sr[sr.groupby(sr.shift(-1).bfill(0).ne(sr)).transform('cumcount')==1].value_counts()

Output:

10    3
12    2
11    2

First column is x value, second is number of sequences.

More compact and faster way:

from  itertools import groupby    
pd.Series([k for k, g in groupby(row)]).value_counts()

Another solution:

np.unique([k for k, g in groupby(row)], return_counts=True)

Result:

(array([10, 11, 12]), array([3, 2, 2], dtype=int64))

Alternatively use np.bincount :

np.bincount([k for k, g in groupby(row)])

But the output will be slightly different:

array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 2, 2], dtype=int64)

Answer 2

I think this is what you want. Groupby the list by similiar elements and then sum the counts

import itertools

element_searched = 10

expected_output = sum([i.count(element_searched) for i in itertools.groupby(list)])

3

Answer 3

I would use a simple dictionary:

row = [10,10,10,10,10,10,10,10,10,11,11,11,11,11,10,10,10,10,12,12,12,12,12,11,11,11,11,12,12,12,12,10]

counter = {}
last_item = None
for item in row:
    if last_item != item:
        counter[item] = counter.get(item, 0) + 1
        last_item = item

print (counter)

Answer 4

A few thoughts which should guide you:

• you need a variable which will store the previous value
• a map of elements, the key being the value of the element, the value being the number of occurrences found so far
• on each iteration check if the current element equals the previous element and if not, then increment the map item having the current element as key
• set previous to current value at each step