如何从嵌套在特定元素之间的列表中提取元素，并将它们添加到新列表中？

Question

I have a daypart column (str), which has 1s or 0s for each hour of the day, depending if we choose to run a campaign during that hour.

Example:

daypart = '110011100111111100011110'

I want to convert this to the following string format:

'0-1, 4-6, 9-15, 19-22'

The above format is more readable, and shows during which hours the campaign ran.

Here's what I'm doing:

hours_list = []

ind = 0
for x in daypart:
    if int(x) == 1:
        hours_list.append(ind)
    else:
        hours_list.append('exclude')
    ind += 1

The above gives me a list like this:

[0, 1, 'exclude', 'exclude', 4, 5, 6, 'exclude', 'exclude', 9, 10, 11, 12, 13, 14, 15, 'exclude', 'exclude', 'exclude', 19, 20, 21, 22, 'exclude']

Now I want to find a way to make the above into my desired output. What I am thinking of doing is finding which elements exist between 'exclude', and start adding them to new lists. I can then take the smallest and largest element from each list, join them with a '-', and append all such lists together.

Any ideas how I can do this, or a simpler way to do all of this?

Answer 1

Here's simple, readable code to get all intervals:

daypart = '1111111111111111111111'
hours= []
start, end = -1, -1
for i in range(len(daypart)):
    if daypart[i] == "1":
        if end != -1:
            end += 1 
        else:
            start = i 
            end = i 
    else:
        if end!=-1:
            hours.append([start, end])
            start, end = -1,-1
if end!=-1:
    hours.append([start, end])
    start, end = -1,-1

print(hours)

Answer 2

I suggest that you convert directly to your desired format rather than using an intermediate representation that has the exact same information as the original input. Let's think about how we can do this in words:

1. Look for the first 1 in the input string
2. Add the index to a list
3. Look for the next 0 in the string.
4. Append one less than found index to a list. (Or maybe append the index from steps 2 and 4 as a pair?)
5. Continue by looking for the next 1 and repeat steps 2-4.

I leave translating this into code as an exercise for the reader.

Answer 3

This can be done using itertools.groupby , operator.itemgetter , enumerate in a comprehension to achieve this as well:

from itertools import groupby
from operator import itemgetter
daypart = '110011100111111100011110'
get_ends, get_one = itemgetter(0,-1), itemgetter(1)

output = ', '.join('{0[0]}-{1[0]}'.format(*get_ends(list(g))) for k,g in groupby(enumerate(daypart), get_one) if k=='1')
print(output)

---

0-1, 4-6, 9-15, 19-22

get_ends gets the first and last elements in each group and get_one just gets element 1 so to use it as a key.