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How to append strings with tab characters to a list and print them also with tab characters?

I want to append a string with a tab character to a list. Then, when I print the list, I want to be able to get the same output when I do the print(string).
For example:
list1 = []
string = "Names:\tLucas"
print(list1)
print(string)
The output:
['Names\tLucas']
Names Lucas
What's a way to do this.

You will have to print the list items themselves, instead of the list.
By iteration:
>>> list1 = []
>>> string1 = "Names:\tLucas"
>>> string2 = "Names:\tJohn"
>>>
>>> list1.append(string1)
>>> list1.append(string2)
>>>
>>> for item in list1:
... print(item)
...
Names: Lucas
Names: John
By unpacking the list:
>>> list1 = []
>>> string1 = "Names:\tLucas"
>>> string2 = "Names:\tJohn"
>>>
>>> list1.append(string1)
>>> list1.append(string2)
>>>
>>> print(*list1, sep='\n')
Names: Lucas
Names: John

try something like this
print("{}".format(string))

Python: Split a list into multiple lists based on a subset of elements [duplicate]

This question already has answers here:
Splitting a list based on a delimiter word
(4 answers)
Closed 4 years ago.
I am trying to split a list that I have into individual lists whenever a specific character or a group of characters occur.
eg.
Main_list = [ 'abcd 1233','cdgfh3738','hryg21','**L**','gdyrhr657','abc31637','**R**','7473hrtfgf'...]
I want to break this list and save it into a sublist whenever I encounter an 'L' or an 'R'
Desired Result:
sublist_1 = ['abcd 1233','cdgfh3738','hryg21']
sublist_2 = ['gdyrhr657','abc31637']
sublist 3 = ['7473hrtfgf'...]
Is there a built in function or a quick way to do this ?
Edit: I do not want the delimiter to be in the list

Use a dictionary for a variable number of variables.
In this case, you can use itertools.groupby to efficiently separate your lists:
L = ['abcd 1233','cdgfh3738','hryg21','**L**',
'gdyrhr657','abc31637','**R**','7473hrtfgf']
from itertools import groupby
# define separator keys
def split_condition(x):
return x in {'**L**', '**R**'}
# define groupby object
grouper = groupby(L, key=split_condition)
# convert to dictionary via enumerate
res = dict(enumerate((list(j) for i, j in grouper if not i), 1))
print(res)
{1: ['abcd 1233', 'cdgfh3738', 'hryg21'],
2: ['gdyrhr657', 'abc31637'],
3: ['7473hrtfgf']}

Consider using one of many helpful tools from a library, i.e. more_itertools.split_at:
Given
import more_itertools as mit
lst = [
"abcd 1233", "cdgfh3738", "hryg21", "**L**",
"gdyrhr657", "abc31637", "**R**",
"7473hrtfgf"
]
Code
result = list(mit.split_at(lst, pred=lambda x: set(x) & {"L", "R"}))
Demo
sublist_1, sublist_2, sublist_3 = result
sublist_1
# ['abcd 1233', 'cdgfh3738', 'hryg21']
sublist_2
# ['gdyrhr657', 'abc31637']
sublist_3
# ['7473hrtfgf']
Details
The more_itertools.split_at function splits an iterable at positions that meet a special condition. The conditional function (predicate) happens to be a lambda function, which is equivalent to and substitutable with the following regular function:
def pred(x):
a = set(x)
b = {"L", "R"}
return a.intersection(b)
Whenever characters of string x intersect with L or R, the predicate returns True, and the split occurs at that position.
Install this package at the commandline via > pip install more_itertools.

#Polyhedronic, you can also try this.
>>> import re
>>> Main_list = [ 'abcd 1233','cdgfh3738','hryg21','**L**','gdyrhr657','abc31637','**R**','7473hrtfgf']
>>>
>>> s = ','.join(Main_list)
>>> s
'abcd 1233,cdgfh3738,hryg21,**L**,gdyrhr657,abc31637,**R**,7473hrtfgf'
>>>
>>> items = re.split('\*\*R\*\*|\*\*L\*\*', s)
>>> items
['abcd 1233,cdgfh3738,hryg21,', ',gdyrhr657,abc31637,', ',7473hrtfgf']
>>>
>>> output = [[a for a in item.split(',') if a] for item in items]
>>> output
[['abcd 1233', 'cdgfh3738', 'hryg21'], ['gdyrhr657', 'abc31637'], ['7473hrtfgf']]
>>>
>>> sublist_1 = output[0]
>>> sublist_2 = output[1]
>>> sublist_3 = output[2]
>>>
>>> sublist_1
['abcd 1233', 'cdgfh3738', 'hryg21']
>>>
>>> sublist_2
['gdyrhr657', 'abc31637']
>>>
>>> sublist_3
['7473hrtfgf']
>>>

Python Convert a part of a String into Float

I am trying to make a list containing 2 strings:
List=["Hight = 7.2", "baselength = 8.32"]
But I am having a problem trying to extract the numbers from the strings:
For example:
If "Hight = 7.2" then the result should be: 7.2
or if the "Hight= 7.3232" then the result should be: 7.3232

Using re.findall :
>>> out = []
>>> for s in l:
out.append( float(re.findall('\d+(?:\.\d+)?', s)[0]) )
>>> out
=> [7.2, 8.0]
Or, without regex, using split,
>>> out = []
>>> for s in l:
num = s.replace(' ','').split('=')[1]
#note : removed whitespace so don't have to deal with cases like
# `n = 2` or `n=2`
out.append(float(num))
>>> out
=> [7.2, 8.0]
#driver values :
IN : l = ["Hight = 7.2","baselength = 8"]

How about this
[(item.split('=')[0],float(item.split('=')[1]) ) for item in List]
Output :
[('Hight ', 7.2), ('baselength ', 8.32)]

Having a label associated to a value is best managed with a dictionary, however if you must have each label=value pair as an entry in a list because perhaps you are reading it into Python from elsewhere, you could use the re module to extract the numeric value from each string in the list:
import re
list=["height = 7.2", "length = 8.32"]
for dim in list:
print(float(re.search('\d+.\d+', dim).group()))

You could convert your list to a dictionary using a comprehension:
import re
List=["Height = 7.2", "baselength = 8.32"]
rx = re.compile(r'(?P<key>\w+)\s*=\s*(?P<value>\d+(?:\.\d+)?)')
Dict = {m.group('key'): float(m.group('value'))
for item in List
for m in [rx.search(item)]}
print(Dict)
# {'Height': 7.2, 'baselength': 8.32}
Afterwards, you can access your values with e.g. Dict["Height"] (here: 7.2).

It's very simple. Use this method for any type of value
List=["Hight = 7.2", "baselength = 8.32"]
# showing example for one value , but you can loop the entire list
a = List[0].split("= ")[1] #accessing first element and split with "= "
print a
'7.2'

Moving parts of string around python

I have a string, well, several actually. The strings are simply:
string.a.is.this
or
string.a.im
in that fashion.
and what I want to do is make those stings become:
this.is.a.string
and
im.a.string
What I've tried:
new_string = string.split('.')
new_string = (new_string[3] + '.' + new_string[2] + '.' + new_string[1] + '.' + new_string[0])
Which works fine for making:
string.a.is.this
into
this.is.a.string
but gives me a error of 'out of range' if I try it on:
string.a.im
yet if I do:
new_string = (new_string[2] + '.' + new_string[1] + '.' + new_string[0])
that works fine to make:
string.a.im
into
im.a.string
but obviously does not work for:
string.a.is.this
since it is not setup for 4 indices. I was trying to figure out how to make the extra index optional, or any other work around, or, better method. Thanks.

You can use str.join, str.split, and [::-1]:
>>> mystr = 'string.a.is.this'
>>> '.'.join(mystr.split('.')[::-1])
'this.is.a.string'
>>> mystr = 'string.a.im'
>>> '.'.join(mystr.split('.')[::-1])
'im.a.string'
>>>
To explain better, here is a step-by-step demonstration with the first string:
>>> mystr = 'string.a.is.this'
>>>
>>> # Split the string on .
>>> mystr.split('.')
['string', 'a', 'is', 'this']
>>>
>>> # Reverse the list returned above
>>> mystr.split('.')[::-1]
['this', 'is', 'a', 'string']
>>>
>>> # Join the strings in the reversed list, separating them by .
>>> '.'.join(mystr.split('.')[::-1])
'this.is.a.string'
>>>

You could do it through python's re module,
import re
mystr = 'string.a.is.this'
regex = re.findall(r'([^.]+)', mystr)
'.'.join(regex[::-1])
'this.is.a.string'

Split string into tuple (Upper,lower) 'ABCDefgh' . Python 2.7.6

my_string = 'ABCDefgh'
desired = ('ABCD','efgh')
the only way I can think of doing this is creating a for loop and then scanning through and checking each element in the string individually and adding to string and then creating the tuple . . . is there a more efficient way to do this?
it will always be in the format UPPERlower

print re.split("([A-Z]+)",my_string)[1:]

Simple way (two passes):
>>> import itertools
>>> my_string = 'ABCDefgh'
>>> desired = (''.join(itertools.takewhile(lambda c:c.isupper(), my_string)), ''.join(itertools.dropwhile(lambda c:c.isupper(), my_string)))
>>> desired
('ABCD', 'efgh')
Efficient way (one pass):
>>> my_string = 'ABCDefgh'
>>> uppers = []
>>> done = False
>>> i = 0
>>> while not done:
... c = my_string[i]
... if c.isupper():
... uppers.append(c)
... i += 1
... else:
... done = True
...
>>> lowers = my_string[i:]
>>> desired = (''.join(uppers), lowers)
>>> desired
('ABCD', 'efgh')

Because I throw itertools.groupby at everything:
>>> my_string = 'ABCDefgh'
>>> from itertools import groupby
>>> [''.join(g) for k,g in groupby(my_string, str.isupper)]
['ABCD', 'efgh']
(A little overpowered here, but scales up to more complicated problems nicely.)

my_string='ABCDefg'
import re
desired = (re.search('[A-Z]+',my_string).group(0),re.search('[a-z]+',my_string).group(0))
print desired

A more robust approach without using re
import string
>>> txt = "ABCeUiioualfjNLkdD"
>>> tup = (''.join([char for char in txt if char in string.ascii_uppercase]),
''.join([char for char in txt if char not in string.ascii_uppercase]))
>>> tup
('ABCUNLD', 'eiioualfjkd')
the char not in string.ascii_uppercase instead of char in string.ascii_lowercase means that you'll never lose any data in case your string has non-letters in it, which could be useful if you suddenly start having errors when this input starts being rejected 20 function calls later.