如何检查 python 列表中连续字符的长度？

Question

Basically say I have a list in python

lst = [6, 6, 7, 1, 1, 1, 1, 4, 1]

the program should return 4 because the number '1' is displayed 4 times in a consecutive manner. Btw the number 6 also displays in a consecutive manner but the program should get the number that shows up the most while being consecutive

more examples:

[6, 4, 4, 4, 1, 1, 7]

should output 3 because '4' is consecutive 3 times

I've been stuck on this problem for a while now, any advice??

Answer 1

groupby() will give you individual groups of the same value by default:

> [list(g) for k, g in groupby(lst)] 
[[6, 6], [7], [1, 1, 1, 1], [4], [1]]

You can pass that to max() with a key of len to get the longest one:

> from itertools import group-by
> lst = [6, 6, 7, 1, 1, 1, 1, 4, 1]
> max((list(g) for k, g in groupby(lst)), key=len)
[1, 1, 1, 1]

That's the longest grouping, from which you can get the length. Alternatively you can just build a list of lengths and call max without the key:

max(len(list(g)) for k, g in groupby(lst))

Answer 2

Two different questions may help.

First to transform the text into list you can check this answer.

Once you have your list, you need to count consecutive values, it was asked and solved here .

Answer 3

Solution

Method-1

You could use list.count() along with set(list) to extract unique values and then enlist the count for each unique value.

lst = [6, 6, 7, 1, 1, 1, 1, 4, 1]
unique = set(lst)
count = [lst.count(k) for k in unique]
print(' unique: {} \n count: {} \n'.format(unique, count))

Output :

 unique: {1, 4, 6, 7} 
 count: [5, 1, 2, 1] 

Method-2

Use numpy.unique to get both the unique values and their respective counts.

import numpy as np

unique, count = np.unique(lst, return_counts=True)
print(' unique: {} \n count: {} \n'.format(unique, count))

Output :

 unique: [1 4 6 7] 
 count: [5 1 2 1] 

Method-3

Use Counter from Colections . This returns a dictionary with the unique values as the keys and the values as their respective counts.

from collections import Counter
Counter(lst)

Output :

Counter({6: 2, 7: 1, 1: 5, 4: 1})