Get the next element of list in Python

Question

I have a list of sentences like:

lst = ['A B C D','E F G H I J','K L M N']

What i did is

l = []
for i in lst:
    for j in i.split():
        print(j)
        l.append(j) 

first = l[::2]
second = l[1::2]

[m+' '+str(n) for m,n in zip(first,second)]

The Output i got is

lst = ['A B', 'C D', 'E F', 'G H', 'I J', 'K L', 'M N']

The Output i want is:

lst = ['A B', 'B C','C D','E F','F G','G H','H I','I J','K L','L M','M N']

I am struggling to think how to achieve this.

[i for x in map(lambda x: x.split(' '), lst) for i in map(' '.join, zip(x[:-1], x[1:]))] — Tarifazo
– Tarifazo, Commented Jul 23, 2019 at 15:43
Your solution is already correct if you replace ::2 with : in both places — Eric
– Eric, Commented Jul 24, 2019 at 16:56

Henry Yik · Accepted Answer · 2019-07-23 15:38:37Z

11

First format your list of string into a list of list, then do a mapping by zip.

i = [i.split() for i in lst]

f = [f"{x} {y}" for item in i for x,y in zip(item,item[1::])]

print (f)

#['A B', 'B C', 'C D', 'E F', 'F G', 'G H', 'H I', 'I J', 'K L', 'L M', 'M N']

edited Jul 23, 2019 at 15:38

answered Jul 23, 2019 at 15:27

Henry Yik

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5 Comments

meowgoesthedog Over a year ago

+1. Can also condense into a one-liner if you replace i with the first list comprehension itself (current code is more readable though)

Tomerikoo Over a year ago

the [x for x in i.split(" ")] could be just i.split()... Nice solution +1

Henry Yik Over a year ago

@Tomerikoo You are right. Force of habit to always write what I split on :)

Tomerikoo Over a year ago

I was referring to the whole list comprehension as split already returns a list and not a generator or something... I believe it makes it a bit more readable

Tomerikoo Over a year ago

Cool, if I could, would be +2 :)

Tomerikoo · Accepted Answer · 2019-07-23 15:26:02Z

Your problem is that you're flattening the whole list and dividing to couples when you want to divide to subsequent couples only the inner elements. So for that we will perform the operation on each element separatly:

lst = ['A B C D','E F G H I J','K L M N']

res = []
for s in lst:
    sub_l = s.split()
    for i in range(len(sub_l)-1):
        res.append("{} {}".format(sub_l[i], sub_l[i+1]))
print(res)

Gives:

['A B', 'B C', 'C D', 'E F', 'F G', 'G H', 'H I', 'I J', 'K L', 'L M', 'M N']

ncica · Accepted Answer · 2019-07-23 15:28:24Z

2

nested = []
for item in lst:
    item = (' '.join(item).split())
    for ix in range(len(item) - 1):
        nested.append(' '.join(item[ix:ix + 2]))

print (nested)

output:

['A B', 'B C', 'C D', 'E F', 'F G', 'G H', 'H I', 'I J', 'K L', 'L M', 'M N']

edited Jul 23, 2019 at 15:28

answered Jul 23, 2019 at 15:22

ncica

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1 Comment

Lucas Wieloch Over a year ago

@Tomerikoo is correct, the output is slightly different, I was doing the same as @ncica. You'll probably need to add some if condition to not append a few elements

Rob Kwasowski · Accepted Answer · 2019-07-24 09:50:27Z

1

Here is a method using Regex:

import re
lst = ['A B C D','E F G H I J','K L M N']
result = re.findall(r'(?=(\b\w+?\b \b\w+?\b))', str(lst))
print(result)

Output:

['A B', 'B C', 'C D', 'E F', 'F G', 'G H', 'H I', 'I J', 'K L', 'L M', 'M N']

edited Jul 24, 2019 at 9:50

answered Jul 23, 2019 at 15:30

Rob Kwasowski

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2 Comments

james joyce Over a year ago

Thnks for the answer...Yes i have already tried it and its not working it gives only the last and first value of both words....every words in list is actual word not an alphabet.

Rob Kwasowski Over a year ago

@jamesjoyce I've edited it so now it will work for words. And please be more specific in your questions in future because you didn't mention that it would be words and not just letters.

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