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join string lists
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Jun-11-2020, 01:37 PM
Hello!
I have a lists which I want to join with the second item of itself for exmaple: 1=[A, B, C, D] I want python to output: 1=[AB, BC, CD] how can I manage that?
Jun-11-2020, 02:06 PM
lst = ['a','b','c','d'] joinlst = ''.join(lst) newlist = [joinlst[:2],joinlst[1:3], joinlst[2:4]] print(newlist)
I welcome all feedback.
The only dumb question, is one that doesn't get asked. My Github How to post code using bbtags Download my project scripts
If you start with:
1=[A, B, C, D] you are going to have a hard time making this work. - You need a legal name for your list - If A,B,C,... are strings , they seem to have no value (or is it 'A'...). We need a better starting point. Paul
It is more important to do the right thing, than to do the thing right.(P.Drucker)
Better is the enemy of good. (Montesquieu) = French version for 'kiss'.
Jun-11-2020, 02:15 PM
(Jun-11-2020, 02:06 PM)DPaul Wrote: If you start with: ok, so lets use this as the staring case: anylist=['A', 'B', 'C', 'D'] i want to output: newlist=['AB', 'BC', 'CD']
Another way
lst = ['a','b','c','d']
joinlst = ''.join(lst)
newlist = []
for i in range(len(joinlst)):
newlist.append(joinlst[i:i+2])
newlist.pop()
print(newlist)
I welcome all feedback.
The only dumb question, is one that doesn't get asked. My Github How to post code using bbtags Download my project scripts
Jun-11-2020, 02:33 PM
I'm curious to hear from some of the more experienced coders on the forum what they think is the best way to do this. I understand that it's generally considered desirable to avoid using
range(len(sequence)), but I'm having a hard time doing so without adding lines of code or checks I wouldn't need otherwise.Here is what I came up with: l = ['a', 'b', 'c', 'd']
newl = []
for i in range(len(l) -1):
newl.append(''.join(l[i] + l[i+1]))
print(newl)
Jun-11-2020, 02:42 PM
There is built-in module textwrap - Text wrapping and filling which have function wrap
>>> import textwrap
>>> lst = ['A', 'B', 'C', 'D']
>>> textwrap.wrap(''.join(lst), 2)
['AB', 'CD']
>>> lst = ['A', 'B', 'C', 'D', 'E']
>>> textwrap.wrap(''.join(lst), 2)
['AB', 'CD', 'E']
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Jun-11-2020, 02:56 PM
(Jun-11-2020, 02:33 PM)GOTO10 Wrote: I'm curious to hear from some of the more experienced coders on the forum what they think is the best way to do this. I understand that it's generally considered desirable to avoid using that worked perfectly for me. thank you
Jun-11-2020, 04:41 PM
(This post was last modified: Jun-11-2020, 04:41 PM by pyzyx3qwerty.)
Not the best approach, but if you wanted, you could do
list1 = ["a","b","c","d"] str1 = list1[0] + list1[1] str2 = list1[1] + list1[2] str3 = list1[2] + list1[3] list2 = [str1,str2,str3] print(list2)And output : list2 and print it, like:print([str1,str2,str3])
pyzyx3qwerty
"The greatest glory in living lies not in never falling, but in rising every time we fall." - Nelson Mandela Need help on the forum? Visit help @ python forum For learning more and more about python, visit Python docs
Assuming that if you only pass a list with one item it returns an empty list.
from string import ascii_uppercase
def adjacent_list(iterabe):
stored_character = None
for character in iterabe:
if stored_character:
yield ''.join((stored_character, character))
stored_character = character
for index in range(1, 27):
letters = (ascii_uppercase[:index])
print(list(adjacent_list(letters))) |
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