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Slice string at last digit in Python

So I have strings with a date somewhere in the middle, like 111_Joe_Smith_2010_Assessment and I want to truncate them such that they become something like 111_Joe_Smith_2010. The code that I thought would work is
reverseString = currentString[::-1]
stripper = re.search('\d', reverseString)
But for some reason this doesn't always give me the right result. Most of the time it does, but every now and then, it will output a string that looks like 111_Joe_Smith_2010_A.
If anyone knows what's wrong with this, it would be super helpful!

You can use re.sub and $ to match and substitute alphabetical characters
and underscores until the end of the string:
import re
d = ['111_Joe_Smith_2010_Assessment', '111_Bob_Smith_2010_Test_assessment']
new_s = [re.sub('[a-zA-Z_]+$', '', i) for i in d]
Output:
['111_Joe_Smith_2010', '111_Bob_Smith_2010']

You could strip non-digit characters from the end of the string using re.sub like this:
>>> import re
>>> re.sub(r'\D+$', '', '111_Joe_Smith_2010_Assessment')
'111_Joe_Smith_2010'
For your input format you could also do it with a simple loop:
>>> s = '111_Joe_Smith_2010_Assessment'
>>> i = len(s) - 1
>>> while not s[i].isdigit():
... i -= 1
...
>>> s[:i+1]
'111_Joe_Smith_2010'

You can use the following approach:
def clean_names():
names = ['111_Joe_Smith_2010_Assessment', '111_Bob_Smith_2010_Test_assessment']
for name in names:
while not name[-1].isdigit():
name = name[:-1]
print(name)

Here is another solution using rstrip() to remove trailing letters and underscores, which I consider a pretty smart alternative to re.sub() as used in other answers:
import string
s = '111_Joe_Smith_2010_Assessment'
new_s = s.rstrip(f'{string.ascii_letters}_') # For Python 3.6+
new_s = s.rstrip(string.ascii_letters+'_') # For other Python versions
print(new_s) # 111_Joe_Smith_2010

split string python with changing indicator

lets say I have a string .
a = '!!!!!!a1######a2&&&&&&a3::::'
It naturally splits by: a1,a2 and a3 to
['!!!!!!','######','&&&&&&','::::']
I want to use the split function, something like:
>>> a.split('a*')
The * indicates that it doesn't matter what character comes after a. Is there an immediate way to do this?

s = '!!!!!!a1######a2&&&&&&a3::::'
import re
print(re.split(r'a[0-9]+', s))

By using regex, with module re, you can try like this:
import re
a = re.split(r'a\d','!!!!!!a1######a2&&&&&&a3::::')
If you want to be more specific in your split keys, try this:
a = re.split(r'a1|a2|a3','!!!!!!a1######a2&&&&&&a3::::')
and make your custom condition, as you wish.

While not as efficient as #Menglong's solution, you could technically do it with just list and string operations, without importing re:
>>> a = '!!!!!!a1######a2&&&&&&a3::::'
>>> s = a.split('a')
>>> s[:1] + [x[1:] for x in s[1:] if x]
['!!!!!!', '######', '&&&&&&', '::::']
This works because if you split on 'a', the first character of every segment after the first will be the * character you want to get rid of.
This solution is not preferable, just something I did as an exercise.

Another alternating-case in-a-string in Python 3.+

I'm very new to Python and am trying to understand how to manipulate strings.
What I want to do is change a string by removing the spaces and alternating the case from upper to lower, IE "This is harder than I thought it would be" to "ThIsIsHaRdErThAnItHoUgHtItWoUlDbE"
I've cobbled together a code to remove the spaces (heavily borrowed from here):
string1 = input("Ask user for something.")
nospace = ""
for a in string1:
if a == " ":
pass
else:
nospace=nospace+a
... but just can't get my head around the caps/lower case part. There are several similar issues on this site and I've tried amending a few of them, with no joy. I realise I need to define a range and iterate through it, but that's where I draw a blank.
for c in nospace[::]:
d = ""
c = nospace[:1].lower()
d = d + c
c = nospace[:1].upper
print d
All I am getting is a column of V's. I'm obviously getting this very wrong. Please can someone advise where? Thanks in advance.

Here is a cutesie way to do this:
>>> s = "This is harder than I thought it would be"
>>> from itertools import cycle
>>> funcs = cycle([str.upper, str.lower])
>>> ''.join(next(funcs)(c) for c in s if c != ' ')
'ThIsIsHaRdErThAnItHoUgHtItWoUlDbE'
>>>
Or, as suggested by Moses in the comments, you can use str.isspace, which will take care of not just a single space ' '
>>> ''.join(next(funcs)(c) for c in s if not c.isspace())
'ThIsIsHaRdErThAnItHoUgHtItWoUlDbE'
This approach only does a single pass on the string. Although, a two-pass method is likely performant enough.
Now, if you were starting with a nospace string, the best way is to convert to some mutable type (e.g. a list) and use slice-assignment notation. It's a little bit inefficient because it builds intermediate data structures, but slicing is fast in Python, so it may be quite performant. You have to ''.join at the end, to bring it back to a string:
>>> nospace
'ThisisharderthanIthoughtitwouldbe'
>>> nospace = list(nospace)
>>> nospace[0::2] = map(str.upper, nospace[0::2])
>>> nospace[1::2] = map(str.lower, nospace[1::2])
>>> ''.join(nospace)
'ThIsIsHaRdErThAnItHoUgHtItWoUlDbE'
>>>

You're trying to do everything at once. Don't. Break your program into steps.
Read the string.
Remove the spaces from the string (as #A.Sherif just demonstrated here)
Go over the string character by character. If the character is in an odd position, convert it to uppercase. Otherwise, convert to lowercase.

So your 2nd loop is where you're breaking it, because the original list isn't being shortened, the c=nospace[:1] grabs the first character of the string and that's the only character that's ever printed. So a solution would be as follows.
string1 = str(input("Ask user for something."))
nospace = ''.join(string1.split(' '))
for i in range(0, len(nospace)):
if i % 2 == 0:
print(nospace[i].upper(), end="")
else:
print(nospace[i].lower(), end="")
Could also replace the if/else statement with a ternary opperator.
for i in range(0, len(nospace)):
print(nospace[i].upper() if (i % 2 == 0) else nospace[i].lower(), end='')
Final way using enumerate as commented about
for i, c in enumerate(nospace):
print(c.upper() if (i % 2 == 0) else c.lower(), end='')

How to move all special characters to the end of the string in Python?

I'm trying to filter all non-alphanumeric characters to the end of the strings. I am having a hard time with the regex since I don't know where the special characters we be. Here are a couple of simple examples.
hello*there*this*is*a*str*ing*with*asterisks
and&this&is&a&str&ing&&with&ampersands&in&i&t
one%mo%refor%good%mea%sure%I%think%you%get%it
How would I go about sliding all the special characters to the end of the string?
Here is what I tried, but I didn't get anything.
re.compile(r'(.+?)(\**)')
r.sub(r'\1\2', string)
Edit:
Expected output for the first string would be:
hellotherethisisastringwithasterisks********

There's no need for regex here. Just use str.isalpha and build up two lists, then join them:
strings = ['hello*there*this*is*a*str*ing*with*asterisks',
'and&this&is&a&str&ing&&with&ampersands&in&i&t',
'one%mo%refor%good%mea%sure%I%think%you%get%it']
for s in strings:
a = []
b = []
for c in s:
if c.isalpha():
a.append(c)
else:
b.append(c)
print(''.join(a+b))
Result:
hellotherethisisastringwithasterisks********
andthisisastringwithampersandsinit&&&&&&&&&&&
onemoreforgoodmeasureIthinkyougetit%%%%%%%%%%
Alternative print() call for Python 3.5 and higher:
print(*a, *b, sep='')

Here is my proposed solution for this with regex:
import re
def move_nonalpha(string,char):
pattern = "\\"+char
char_list = re.findall(pattern,string)
if len(char_list)>0:
items = re.split(pattern,string)
if len(items)>0:
return ''.join(items)+''.join(char_list)
Usage:
string = "hello*there*this*is*a*str*ing*with*asterisks"
print (move_nonalpha(string,"*"))
Gives me output:
hellotherethisisastringwithasterisks********
I tried with your other input patterns as well and it's working. Hope it'll help.

Python: Dividing a string into substrings

I have a bunch of mathematical expressions stored as strings. Here's a short one:
stringy = "((2+2)-(3+5)-6)"
I want to break this string up into a list that contains ONLY the information in each "sub-parenthetical phrase" (I'm sure there's a better way to phrase that.) So my yield would be:
['2+2','3+5']
I have a couple of ideas about how to do this, but I keep running into a "okay, now what" issue.
For example:
for x in stringy:
substring = stringy[stringy.find('('+1 : stringy.find(')')+1]
stringlist.append(substring)
Works just peachy to return 2+2, but that's about as far as it goes, and I am completely blanking on how to move through the remainder...

One way using regex:
import re
stringy = "((2+2)-(3+5)-6)"
for exp in re.findall("\(([\s\d+*/-]+)\)", stringy):
print exp
Output
2+2
3+5

You could use regular expressions like the following:
import re
x = "((2+2)-(3+5)-6)"
re.findall(r"(?<=\()[0-9+/*-]+(?=\))", x)
Result:
['2+2', '3+5']