Pythonesque way to compare adjacent elements in a list

Question

I am pretty sure this question has been asked about a gazillion times, but I can not find a satisfying answer. I am trying to iterate through a list and find how many occurrences of 'B' are immediately preceded by 'A'. I have a solution, but I am sure it is far from perfect. In C++ I'd do something like (with about 10 variations):

int main()
{
    vector<char> charList={'A', 'B', 'D', 'B', 'C', 'A', 'B'};
    int count=0;
    char prevElem = '\0';
    for(auto x: charList)
    {
        if( x == 'B' && prevElem =='A')
            ++count;
        prevElem = x;
    }
    
    cout << "Count = " << count << endl;

    return 0;
}

What is the right way to do it in Python? I mean the simplest solution is obvious, but what should I do if the data I have is in the form of iterator with the lazy iterable under it and I do not want to go over that iterator twice?

Yes, that is what I mean by "far from perfect". My question is a bit more involved: What should I do if the data I have is in the form of iterator with the lazy iterable under it and I do not want to go over that iterator twice? — Mark Z, Jan 25 '21 at 01:12
Python >= 3.8: `prev = object(); count = sum(1 for char in char_list if (prev, (prev := char)) == ('A', 'B'))`. — ekhumoro, Jan 25 '21 at 01:46
A good place to start for problems like this is the [itertools recipes](https://docs.python.org/3/library/itertools.html#itertools-recipes), in this case `pairwise`. (`sum(map(lambda cc:cc==('A', 'B'), pairwise(charlist)))`). Unlike solutions using subscripting or slicing, this will work on an arbitrary sequence, including generators. — rici, Jan 25 '21 at 01:52
@rici `map(lambda)` is ugly. A generator expression is nicer and shorter: `sum(cc==('A', 'B') for cc in pairwise(charlist))` — wjandrea, Jan 25 '21 at 02:12
I guess the solution by @wjandrea above is what I was looking for! — Mark Z, Jan 25 '21 at 02:25

Robot Jung · Answer 1 · 2021-01-25T01:38:16.443

0

Try this code:

char_list = ['A', 'B', 'D', 'B', 'C', 'A', 'B']

count = 0
for index in range(0, len(char_list)-1):
    if char_list[index] == 'A' and char_list[index+1] == 'B':
        count += 1

print('count :', count)

Try this new one:

char_list = ['A', 'B', 'D', 'B', 'C', 'A', 'B', 'A', 'B']
prev_char = ''

count = 0
for char in char_list:
    if char == 'B' and prev_char == 'A':
        count += 1
    prev_char = char

print('count :', count)

edited Jan 25 '21 at 01:38

answered Jan 25 '21 at 01:02

Robot Jung

367
4
13

That is about what I have now. My question is a bit more involved: What should I do if the data I have is in the form of iterator with the lazy iterable under it and I do not want to go over that iterator twice? – Mark Z Jan 25 '21 at 01:09
@MarkZ You really should have put this information in the question. – Reti43 Jan 25 '21 at 01:11
This would throw an IndexError if the list ends in an `'A'`. – khelwood Jan 25 '21 at 01:12
1

@MarkZ I added a new answer, please review it. – Robot Jung Jan 25 '21 at 01:43
Yep, that is what I was looking for. I went 1 step further. See my own answer too – Mark Z Jan 25 '21 at 01:57

ahmadjanan · Answer 2 · 2021-01-25T01:13:10.157

0

The simplest implementation would be the following:

array = ['A', 'B', 'D', 'B', 'C', 'A', 'B']

count = 0
for i in range(len(array)-1):
  if array[i] == 'A' and array[i+1] == 'B':
    count += 1

print(count)

edited Jan 25 '21 at 01:13

answered Jan 25 '21 at 01:05

ahmadjanan

298
2
9

This would throw an IndexError. – khelwood Jan 25 '21 at 01:06
No, it won't. Because the loop runs 1 time less than the upper-limit. And since the indexing starts from 0 to `len(array) - 1`, it will work just fine. – ahmadjanan Jan 25 '21 at 01:08
1

`i` runs up to `len(array)-1`, so `i+1` runs up to `len(array)`, which will raise an IndexError if the array ends with a `'A'`. – khelwood Jan 25 '21 at 01:11
Ahhh, I see it now. Good catch. I edited the post. – ahmadjanan Jan 25 '21 at 01:13

score 0 · Answer 3 · answered Jan 25 '21 at 01:11

Previus solution will give an error for a list ending in A. I would suggest to start from index 1 and look at the previous one

char_list = ['A', 'B', 'D', 'B', 'C', 'A', 'B']

count = 0
for index in range(1, len(char_list)):
    if char_list[index] == 'B' and char_list[index-1] == 'A':
        count += 1

print('count :', count)

Mark Z · Answer 4 · 2021-01-25T02:14:23.573

0

My version of Robot Jung's answer:

def findAb(data):
    iterator = iter(data)
    prevElem = ''

    for curElem in iterator:
        yield int(curElem == 'B' and prevElem == 'A')
        prevElem = curElem

if __name__ == '__main__':
    data =  ['A', 'B', 'D', 'B', 'C', 'A', 'B']

    count = sum( d for d in findAb(data))
    print (count)

edited Jan 25 '21 at 02:14

answered Jan 25 '21 at 02:04

Mark Z

19
2

How is this an improvement over the much simpler and clearer solution given by RobertJung (i.e. the second one)? – ekhumoro Jan 25 '21 at 02:11
I cleaned up mine slightly, please check again. Whether that is an improvement I do not know. Still learning. I guess it just makes the intent slightly more visible to the user – Mark Z Jan 25 '21 at 02:18
1

On the new version: why not just increment the count inside `findAb` and return it? There's no need for a generator function here. – ekhumoro Jan 25 '21 at 02:21
Probably you are right. Too many available tools – Mark Z Jan 25 '21 at 02:35

Pythonesque way to compare adjacent elements in a list

4 Answers4