Xjyuk

Xcode 10.3 Installation

What gave NASA the confidence for a translunar injection in Apollo 8?

What is the minimum altitude over Egmond aan Zee when landing at Schiphol?

How can I show that the speed of light in vacuum is the same in all reference frames?

Bounded Torsion, without Mazur’s Theorem

How does mathematics work?

Can't understand how static works exactly

Impact of throwing away fruit waste on a peak > 3200 m above a glacier

If hash functions append the length, why does length extension attack work?

What kind of world would drive brains to evolve high-throughput sensory?

Do I care if the housing market has gone up or down, if I'm moving from one house to another?

Revolutionaries' fleet armed with kinetic weapons defeats government fleet armed with missiles

how to add 1 milliseconds on a datetime string?

What is a "staved" town, like in "Staverton"?

How often should alkaline batteries be checked when they are in a device?

"It is what it is" in French

How am I supposed to put out fires?

What is "It is x o'clock" in Japanese with subject

Considerations when providing money to one child now, and the other later?

What does the following chess proverb mean: "Chess is a sea where a gnat may drink from and an elephant may bathe in."

Can a creature sustain itself by eating its own severed body parts?

How can Kazakhstan perform MITM attacks on all HTTPS traffic?

Short story where a flexible reality hardens to an unchanging one

What is the best word describing the nature of expiring in a short amount of time, connoting "losing public attention"?

Get the next element of list in Python

How do I check if a list is empty?What are metaclasses in Python?Finding the index of an item given a list containing it in PythonHow do you split a list into evenly sized chunks?Does Python have a ternary conditional operator?How to print without newline or space?Convert bytes to a string?Getting the last element of a listHow to make a flat list out of list of listsHow do I get the number of elements in a list?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

6

1

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.

python list

share|improve this question

asked 8 hours ago

james joycejames joyce

8477 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  [i for x in map(lambda x: x.split(' '), lst) for i in map(' '.join, zip(x[:-1], x[1:]))]
  
  – Mstaino
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Did you try my answer? It's shorter than the accepted one
  
  – Rob Kwasowski
  3 hours ago
  

  
  
  
  

add a comment | 

6

1

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.

python list

share|improve this question

asked 8 hours ago

james joycejames joyce

8477 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  [i for x in map(lambda x: x.split(' '), lst) for i in map(' '.join, zip(x[:-1], x[1:]))]
  
  – Mstaino
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Did you try my answer? It's shorter than the accepted one
  
  – Rob Kwasowski
  3 hours ago
  

  
  
  
  

add a comment | 

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.

python list

share|improve this question

asked 8 hours ago

james joycejames joyce

8477 bronze badges

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

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)]

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

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

share|improve this question

asked 8 hours ago

james joycejames joyce

8477 bronze badges

share|improve this question

asked 8 hours ago

james joycejames joyce

8477 bronze badges

asked 8 hours ago

james joycejames joyce

8477 bronze badges

asked 8 hours ago

james joycejames joyce

8477 bronze badges

4 Answers
4

active

oldest

votes

6

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']

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Henry YikHenry Yik

3,74822 gold badges55 silver badges2222 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  the [x for x in i.split(" ")] could be just i.split()... Nice solution +1
  
  – Tomerikoo
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Tomerikoo You are right. Force of habit to always write what I split on :)
  
  – Henry Yik
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  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
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Right you are. Don't remember why I ever had to do it - edited.
  
  – Henry Yik
  7 hours ago
  

  
  
  
  

 | 
show 1 more comment

3

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']

share|improve this answer

answered 8 hours ago

TomerikooTomerikoo

2,00877 silver badges1818 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks I got it.. I ran the code line by line.. Thx for explanation..
  
  – james joyce
  7 hours ago
  

  
  
  
  

add a comment | 

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']

share|improve this answer

edited 8 hours ago

answered 8 hours ago

ncicancica

3,18711 gold badge55 silver badges2525 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  
  – Lucas Wieloch
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I edit my post ;)
  
  – ncica
  8 hours ago
  

  
  
  
  

add a comment | 

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('(?=(w w))', 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']

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Rob KwasowskiRob Kwasowski

1,06222 gold badges33 silver badges1919 bronze badges

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f57167309%2fget-the-next-element-of-list-in-python%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

6

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']

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Henry YikHenry Yik

3,74822 gold badges55 silver badges2222 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  the [x for x in i.split(" ")] could be just i.split()... Nice solution +1
  
  – Tomerikoo
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Tomerikoo You are right. Force of habit to always write what I split on :)
  
  – Henry Yik
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  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
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Right you are. Don't remember why I ever had to do it - edited.
  
  – Henry Yik
  7 hours ago
  

  
  
  
  

 | 
show 1 more comment

6

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']

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Henry YikHenry Yik

3,74822 gold badges55 silver badges2222 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  the [x for x in i.split(" ")] could be just i.split()... Nice solution +1
  
  – Tomerikoo
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Tomerikoo You are right. Force of habit to always write what I split on :)
  
  – Henry Yik
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  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
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Right you are. Don't remember why I ever had to do it - edited.
  
  – Henry Yik
  7 hours ago
  

  
  
  
  

 | 
show 1 more comment

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']

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Henry YikHenry Yik

3,74822 gold badges55 silver badges2222 bronze badges

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']

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Henry YikHenry Yik

3,74822 gold badges55 silver badges2222 bronze badges

answered 8 hours ago

Henry YikHenry Yik

3,74822 gold badges55 silver badges2222 bronze badges

answered 8 hours ago

Henry YikHenry Yik

3,74822 gold badges55 silver badges2222 bronze badges

3

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']

share|improve this answer

answered 8 hours ago

TomerikooTomerikoo

2,00877 silver badges1818 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks I got it.. I ran the code line by line.. Thx for explanation..
  
  – james joyce
  7 hours ago
  

  
  
  
  

add a comment | 

3

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']

share|improve this answer

answered 8 hours ago

TomerikooTomerikoo

2,00877 silver badges1818 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks I got it.. I ran the code line by line.. Thx for explanation..
  
  – james joyce
  7 hours ago
  

  
  
  
  

add a comment | 

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']

share|improve this answer

answered 8 hours ago

TomerikooTomerikoo

2,00877 silver badges1818 bronze badges

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)

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

share|improve this answer

answered 8 hours ago

TomerikooTomerikoo

2,00877 silver badges1818 bronze badges

answered 8 hours ago

TomerikooTomerikoo

2,00877 silver badges1818 bronze badges

answered 8 hours ago

TomerikooTomerikoo

2,00877 silver badges1818 bronze badges

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']

share|improve this answer

edited 8 hours ago

answered 8 hours ago

ncicancica

3,18711 gold badge55 silver badges2525 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  
  – Lucas Wieloch
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I edit my post ;)
  
  – ncica
  8 hours ago
  

  
  
  
  

add a comment | 

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']

share|improve this answer

edited 8 hours ago

answered 8 hours ago

ncicancica

3,18711 gold badge55 silver badges2525 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  
  – Lucas Wieloch
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I edit my post ;)
  
  – ncica
  8 hours ago
  

  
  
  
  

add a comment | 

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']

share|improve this answer

edited 8 hours ago

answered 8 hours ago

ncicancica

3,18711 gold badge55 silver badges2525 bronze badges

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)

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

share|improve this answer

edited 8 hours ago

answered 8 hours ago

ncicancica

3,18711 gold badge55 silver badges2525 bronze badges

answered 8 hours ago

ncicancica

3,18711 gold badge55 silver badges2525 bronze badges

answered 8 hours ago

ncicancica

3,18711 gold badge55 silver badges2525 bronze badges

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('(?=(w w))', 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']

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Rob KwasowskiRob Kwasowski

1,06222 gold badges33 silver badges1919 bronze badges

add a comment | 

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('(?=(w w))', 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']

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Rob KwasowskiRob Kwasowski

1,06222 gold badges33 silver badges1919 bronze badges

add a comment | 

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('(?=(w w))', 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']

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Rob KwasowskiRob Kwasowski

1,06222 gold badges33 silver badges1919 bronze badges

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

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

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Rob KwasowskiRob Kwasowski

1,06222 gold badges33 silver badges1919 bronze badges

answered 8 hours ago

Rob KwasowskiRob Kwasowski

1,06222 gold badges33 silver badges1919 bronze badges

answered 8 hours ago

Rob KwasowskiRob Kwasowski

1,06222 gold badges33 silver badges1919 bronze badges