Say I have a string in alphabetical order, based on the amount of times that a letter repeats.
Example: "BBBAADDC".
There are 3 B's, so they go at the start, 2 A's and 2 D's, so the A's go in front of the D's because they are in alphabetical order, and 1 C. Another example would be CCCCAAABBDDAB.
Note that there can be 4 letters in the middle somewhere (i.e. CCCC), as there could be 2 pairs of 2 letters.
However, let's say I can only have n letters in a row. For example, if n = 3 in the second example, then I would have to omit one "C" from the first substring of 4 C's, because there can only be a maximum of 3 of the same letters in a row.
Another example would be the string "CCCDDDAABC"; if n = 2, I would have to remove one C and one D to get the string CCDDAABC
Example input/output:
n=2: Input: AAABBCCCCDE, Output: AABBCCDE
n=4: Input: EEEEEFFFFGGG, Output: EEEEFFFFGGG
n=1: Input: XXYYZZ, Output: XYZ
How can I do this with Python? Thanks in advance!
This is what I have right now, although I'm not sure if it's on the right track. Here, z is the length of the string.
for k in range(z+1):
if final_string[k] == final_string[k+1] == final_string[k+2] == final_string[k+3]:
final_string = final_string.translate({ord(final_string[k]): None})
return final_string
Ok, based on your comment, you're either pre-sorting the string or it doesn't need to be sorted by the function you're trying to create. You can do this more easily with itertools.groupby():
import itertools
def max_seq(text, n=1):
result = []
for k, g in itertools.groupby(text):
result.extend(list(g)[:n])
return ''.join(result)
max_seq('AAABBCCCCDE', 2)
# 'AABBCCDE'
max_seq('EEEEEFFFFGGG', 4)
# 'EEEEFFFFGGG'
max_seq('XXYYZZ')
# 'XYZ'
max_seq('CCCDDDAABC', 2)
# 'CCDDAABC'
In each group g, it's expanded and then sliced until n elements (the [:n] part) so you get each letter at most n times in a row. If the same letter appears elsewhere, it's treated as an independent sequence when counting n in a row.
Edit: Here's a shorter version, which may also perform better for very long strings. And while we're using itertools, this one additionally utilises itertools.chain.from_iterable() to create the flattened list of letters. And since each of these is a generator, it's only evaluated/expanded at the last line:
import itertools
def max_seq(text, n=1):
sequences = (list(g)[:n] for _, g in itertools.groupby(text))
letters = itertools.chain.from_iterable(sequences)
return ''.join(letters)
hello = "hello frrriend"
def replacing() -> str:
global hello
j = 0
for i in hello:
if j == 0:
pass
else:
if i == prev:
hello = hello.replace(i, "")
prev = i
prev = i
j += 1
return hello
replacing()
looks a bit primal but i think it works, thats what i came up with on the go anyways , hope it helps :D
Here's my solution:
def snip_string(string, n):
list_string = list(string)
list_string.sort()
chars = set(string)
for char in chars:
while list_string.count(char) > n:
list_string.remove(char)
return ''.join(list_string)
Calling the function with various values for n gives the following output:
>>> string = "AAAABBBCCCDDD"
>>> snip_string(string, 1)
'ABCD'
>>> snip_string(string, 2)
'AABBCCDD'
>>> snip_string(string, 3)
'AAABBBCCCDDD'
>>>
Edit
Here is the updated version of my solution, which only removes characters if the group of repeated characters exceeds n.
import itertools
def snip_string(string, n):
groups = [list(g) for k, g in itertools.groupby(string)]
string_list = []
for group in groups:
while len(group) > n:
del group[-1]
string_list.extend(group)
return ''.join(string_list)
Output:
>>> string = "DDDAABBBBCCABCDE"
>>> snip_string(string, 3)
'DDDAABBBCCABCDE'
from itertools import groupby
n = 2
def rem(string):
out = "".join(["".join(list(g)[:n]) for _, g in groupby(string)])
print(out)
So this is the entire code for your question.
s = "AABBCCDDEEE"
s2 = "AAAABBBDDDDDDD"
s3 = "CCCCAAABBDDABBB"
s4 = "AAAAAAAA"
z = "AAABBCCCCDE"
With following test:
AABBCCDDEE
AABBDD
CCAABBDDABB
AA
AABBCCDE
Say I have a string 'Area51' and an array ['0051'], how would I go about replacing the 51 in the string with the array so that the output reads 'Area0051'. Assume that I have another function that finds my transform_array but it's not significant to this code.
string = 'Area51'
transformation_array = ['0051']
Ideally, this would extend to examples such as:
string = '22Area51'
transform_array = ['0022','0051']
# Outputting -> '0022Area0051'
I know strings are immutable so I have to create a new string and can't use replace.
I was thinking something along the lines of:
import re
string = '22Area51'
nums = re.findall("(\d+",string)
transform_array = ['0022','0051']
new_string = ''
for i in range(len(nums)):
k = s.index(nums[i])
new_string += string[s[:k] + transform_array[i]
But this would output:
First iteration:
>>> '0022Area51'
Second iteration
>>> '22Area0051'
I can't seem to wrap my mind on how to put it together. Any guidance would be greatly appreciated.
You can use itertools.cycle (doc) and re.sub with custom sub function:
string = '22Area51'
transform_array = ['0022','0051']
import re
from itertools import cycle
c = cycle(transform_array)
print(re.sub(r'\d+', lambda g: next(c), string))
Prints:
0022Area0051
Or, if number of digit groups matches the length of transform array:
import re
c = iter(transform_array)
print(re.sub(r'\d+', lambda g: next(c), string))
With simple builtin iter feature:
import re
string = '22Area51'
transform_array = ['0022','0051']
tr_arr_iter = iter(transform_array) # prepare iterator
res = re.sub(r'\d+', lambda n: next(tr_arr_iter), string)
print(res) # 0022Area0051
import re
string = '22Area51'
transform_array = ['0022', '0051']
new_string = string
nums = re.findall(r'\d+', string)
for num in nums:
for el in transform_array:
if num in el:
new_string = new_string.replace(num, el)
print(new_string) #0022Area0051
how can i write a python program to intake some alphabets in and print out (alphabets+n) in the output. Example
my_string = 'abc'
expected_output = 'cde' # n=2
One way I've thought is by using str.maketrans, and mapping the original input to (alphabets + n). Is there any other way?
PS: xyz should translate to abc
I've tried to write my own code as well for this, (apart from the infinitely better answers mentioned):
number = 2
prim = """abc! fgdf """
final = prim.lower()
for x in final:
if(x =="y"):
print("a", end="")
elif(x=="z"):
print("b", end="")
else:
conv = ord(x)
x = conv+number
print(chr(x),end="")
Any comments on how to not convert special chars? thanks
If you don't care about wrapping around, you can just do:
def shiftString(string, number):
return "".join(map(lambda x: chr(ord(x)+number),string))
If you do want to wrap around (think Caesar chiffre), you'll need to specify a start and an end of where the alphabet begins and ends:
def shiftString(string, number, start=97, num_of_symbols=26):
return "".join(map(lambda x: chr(((ord(x)+number-start) %
num_of_symbols)+start) if start <= ord(x) <= start+num_of_symbols
else x,string))
That would, e.g., convert abcxyz, when given a shift of 2, into cdezab.
If you actually want to use it for "encryption", make sure to exclude non-alphabetic characters (like spaces etc.) from it.
edit: Shameless plug of my Vignère tool in Python
edit2: Now only converts in its range.
How about something like
>>> my_string = "abc"
>>> n = 2
>>> "".join([ chr(ord(i) + n) for i in my_string])
'cde'
Note As mentioned in comments the question is bit vague about what to do when the edge cases are encoundered like xyz
Edit To take care of edge cases, you can write something like
>>> from string import ascii_lowercase
>>> lower = ascii_lowercase
>>> input = "xyz"
>>> "".join([ lower[(lower.index(i)+2)%26] for i in input ])
'zab'
>>> input = "abc"
>>> "".join([ lower[(lower.index(i)+2)%26] for i in input ])
'cde'
I've made the following change to the code:
number = 2
prim = """Special() ops() chars!!"""
final = prim.lower()
for x in final:
if(x =="y"):
print("a", end="")
elif(x=="z"):
print("b", end="")
elif (ord(x) in range(97, 124)):
conv = ord(x)
x = conv+number
print(chr(x),end="")
else:
print(x, end="")
**Output**: urgekcn() qru() ejctu!!
test_data = (('abz', 2), ('abc', 3), ('aek', 26), ('abcd', 25))
# translate every character
def shiftstr(s, k):
if not (isinstance(s, str) and isinstance(k, int) and k >=0):
return s
a = ord('a')
return ''.join([chr(a+((ord(c)-a+k)%26)) for c in s])
for s, k in test_data:
print(shiftstr(s, k))
print('----')
# translate at most 26 characters, rest look up dictionary at O(1)
def shiftstr(s, k):
if not (isinstance(s, str) and isinstance(k, int) and k >=0):
return s
a = ord('a')
d = {}
l = []
for c in s:
v = d.get(c)
if v is None:
v = chr(a+((ord(c)-a+k)%26))
d[c] = v
l.append(v)
return ''.join(l)
for s, k in test_data:
print(shiftstr(s, k))
Testing shiftstr_test.py (above code):
$ python3 shiftstr_test.py
cdb
def
aek
zabc
----
cdb
def
aek
zabc
It covers wrapping.
Python has string.find() and string.rfind() to get the index of a substring in a string.
I'm wondering whether there is something like string.find_all() which can return all found indexes (not only the first from the beginning or the first from the end).
For example:
string = "test test test test"
print string.find('test') # 0
print string.rfind('test') # 15
#this is the goal
print string.find_all('test') # [0,5,10,15]
For counting the occurrences, see Count number of occurrences of a substring in a string.
There is no simple built-in string function that does what you're looking for, but you could use the more powerful regular expressions:
import re
[m.start() for m in re.finditer('test', 'test test test test')]
#[0, 5, 10, 15]
If you want to find overlapping matches, lookahead will do that:
[m.start() for m in re.finditer('(?=tt)', 'ttt')]
#[0, 1]
If you want a reverse find-all without overlaps, you can combine positive and negative lookahead into an expression like this:
search = 'tt'
[m.start() for m in re.finditer('(?=%s)(?!.{1,%d}%s)' % (search, len(search)-1, search), 'ttt')]
#[1]
re.finditer returns a generator, so you could change the [] in the above to () to get a generator instead of a list which will be more efficient if you're only iterating through the results once.
>>> help(str.find)
Help on method_descriptor:
find(...)
S.find(sub [,start [,end]]) -> int
Thus, we can build it ourselves:
def find_all(a_str, sub):
start = 0
while True:
start = a_str.find(sub, start)
if start == -1: return
yield start
start += len(sub) # use start += 1 to find overlapping matches
list(find_all('spam spam spam spam', 'spam')) # [0, 5, 10, 15]
No temporary strings or regexes required.
Here's a (very inefficient) way to get all (i.e. even overlapping) matches:
>>> string = "test test test test"
>>> [i for i in range(len(string)) if string.startswith('test', i)]
[0, 5, 10, 15]
Use re.finditer:
import re
sentence = input("Give me a sentence ")
word = input("What word would you like to find ")
for match in re.finditer(word, sentence):
print (match.start(), match.end())
For word = "this" and sentence = "this is a sentence this this" this will yield the output:
(0, 4)
(19, 23)
(24, 28)
Again, old thread, but here's my solution using a generator and plain str.find.
def findall(p, s):
'''Yields all the positions of
the pattern p in the string s.'''
i = s.find(p)
while i != -1:
yield i
i = s.find(p, i+1)
Example
x = 'banananassantana'
[(i, x[i:i+2]) for i in findall('na', x)]
returns
[(2, 'na'), (4, 'na'), (6, 'na'), (14, 'na')]
You can use re.finditer() for non-overlapping matches.
>>> import re
>>> aString = 'this is a string where the substring "is" is repeated several times'
>>> print [(a.start(), a.end()) for a in list(re.finditer('is', aString))]
[(2, 4), (5, 7), (38, 40), (42, 44)]
but won't work for:
In [1]: aString="ababa"
In [2]: print [(a.start(), a.end()) for a in list(re.finditer('aba', aString))]
Output: [(0, 3)]
Come, let us recurse together.
def locations_of_substring(string, substring):
"""Return a list of locations of a substring."""
substring_length = len(substring)
def recurse(locations_found, start):
location = string.find(substring, start)
if location != -1:
return recurse(locations_found + [location], location+substring_length)
else:
return locations_found
return recurse([], 0)
print(locations_of_substring('this is a test for finding this and this', 'this'))
# prints [0, 27, 36]
No need for regular expressions this way.
If you're just looking for a single character, this would work:
string = "dooobiedoobiedoobie"
match = 'o'
reduce(lambda count, char: count + 1 if char == match else count, string, 0)
# produces 7
Also,
string = "test test test test"
match = "test"
len(string.split(match)) - 1
# produces 4
My hunch is that neither of these (especially #2) is terribly performant.
this is an old thread but i got interested and wanted to share my solution.
def find_all(a_string, sub):
result = []
k = 0
while k < len(a_string):
k = a_string.find(sub, k)
if k == -1:
return result
else:
result.append(k)
k += 1 #change to k += len(sub) to not search overlapping results
return result
It should return a list of positions where the substring was found.
Please comment if you see an error or room for improvment.
This does the trick for me using re.finditer
import re
text = 'This is sample text to test if this pythonic '\
'program can serve as an indexing platform for '\
'finding words in a paragraph. It can give '\
'values as to where the word is located with the '\
'different examples as stated'
# find all occurances of the word 'as' in the above text
find_the_word = re.finditer('as', text)
for match in find_the_word:
print('start {}, end {}, search string \'{}\''.
format(match.start(), match.end(), match.group()))
This thread is a little old but this worked for me:
numberString = "onetwothreefourfivesixseveneightninefiveten"
testString = "five"
marker = 0
while marker < len(numberString):
try:
print(numberString.index("five",marker))
marker = numberString.index("five", marker) + 1
except ValueError:
print("String not found")
marker = len(numberString)
You can try :
>>> string = "test test test test"
>>> for index,value in enumerate(string):
if string[index:index+(len("test"))] == "test":
print index
0
5
10
15
You can try :
import re
str1 = "This dress looks good; you have good taste in clothes."
substr = "good"
result = [_.start() for _ in re.finditer(substr, str1)]
# result = [17, 32]
When looking for a large amount of key words in a document, use flashtext
from flashtext import KeywordProcessor
words = ['test', 'exam', 'quiz']
txt = 'this is a test'
kwp = KeywordProcessor()
kwp.add_keywords_from_list(words)
result = kwp.extract_keywords(txt, span_info=True)
Flashtext runs faster than regex on large list of search words.
This function does not look at all positions inside the string, it does not waste compute resources. My try:
def findAll(string,word):
all_positions=[]
next_pos=-1
while True:
next_pos=string.find(word,next_pos+1)
if(next_pos<0):
break
all_positions.append(next_pos)
return all_positions
to use it call it like this:
result=findAll('this word is a big word man how many words are there?','word')
src = input() # we will find substring in this string
sub = input() # substring
res = []
pos = src.find(sub)
while pos != -1:
res.append(pos)
pos = src.find(sub, pos + 1)
Whatever the solutions provided by others are completely based on the available method find() or any available methods.
What is the core basic algorithm to find all the occurrences of a
substring in a string?
def find_all(string,substring):
"""
Function: Returning all the index of substring in a string
Arguments: String and the search string
Return:Returning a list
"""
length = len(substring)
c=0
indexes = []
while c < len(string):
if string[c:c+length] == substring:
indexes.append(c)
c=c+1
return indexes
You can also inherit str class to new class and can use this function
below.
class newstr(str):
def find_all(string,substring):
"""
Function: Returning all the index of substring in a string
Arguments: String and the search string
Return:Returning a list
"""
length = len(substring)
c=0
indexes = []
while c < len(string):
if string[c:c+length] == substring:
indexes.append(c)
c=c+1
return indexes
Calling the method
newstr.find_all('Do you find this answer helpful? then upvote
this!','this')
This is solution of a similar question from hackerrank. I hope this could help you.
import re
a = input()
b = input()
if b not in a:
print((-1,-1))
else:
#create two list as
start_indc = [m.start() for m in re.finditer('(?=' + b + ')', a)]
for i in range(len(start_indc)):
print((start_indc[i], start_indc[i]+len(b)-1))
Output:
aaadaa
aa
(0, 1)
(1, 2)
(4, 5)
Here's a solution that I came up with, using assignment expression (new feature since Python 3.8):
string = "test test test test"
phrase = "test"
start = -1
result = [(start := string.find(phrase, start + 1)) for _ in range(string.count(phrase))]
Output:
[0, 5, 10, 15]
I think the most clean way of solution is without libraries and yields:
def find_all_occurrences(string, sub):
index_of_occurrences = []
current_index = 0
while True:
current_index = string.find(sub, current_index)
if current_index == -1:
return index_of_occurrences
else:
index_of_occurrences.append(current_index)
current_index += len(sub)
find_all_occurrences(string, substr)
Note: find() method returns -1 when it can't find anything
The pythonic way would be:
mystring = 'Hello World, this should work!'
find_all = lambda c,s: [x for x in range(c.find(s), len(c)) if c[x] == s]
# s represents the search string
# c represents the character string
find_all(mystring,'o') # will return all positions of 'o'
[4, 7, 20, 26]
>>>
if you only want to use numpy here is a solution
import numpy as np
S= "test test test test"
S2 = 'test'
inds = np.cumsum([len(k)+len(S2) for k in S.split(S2)[:-1]])- len(S2)
print(inds)
if you want to use without re(regex) then:
find_all = lambda _str,_w : [ i for i in range(len(_str)) if _str.startswith(_w,i) ]
string = "test test test test"
print( find_all(string, 'test') ) # >>> [0, 5, 10, 15]
please look at below code
#!/usr/bin/env python
# coding:utf-8
'''黄哥Python'''
def get_substring_indices(text, s):
result = [i for i in range(len(text)) if text.startswith(s, i)]
return result
if __name__ == '__main__':
text = "How much wood would a wood chuck chuck if a wood chuck could chuck wood?"
s = 'wood'
print get_substring_indices(text, s)
def find_index(string, let):
enumerated = [place for place, letter in enumerate(string) if letter == let]
return enumerated
for example :
find_index("hey doode find d", "d")
returns:
[4, 7, 13, 15]
Not exactly what OP asked but you could also use the split function to get a list of where all the substrings don't occur. OP didn't specify the end goal of the code but if your goal is to remove the substrings anyways then this could be a simple one-liner. There are probably more efficient ways to do this with larger strings; regular expressions would be preferable in that case
# Extract all non-substrings
s = "an-example-string"
s_no_dash = s.split('-')
# >>> s_no_dash
# ['an', 'example', 'string']
# Or extract and join them into a sentence
s_no_dash2 = ' '.join(s.split('-'))
# >>> s_no_dash2
# 'an example string'
Did a brief skim of other answers so apologies if this is already up there.
def count_substring(string, sub_string):
c=0
for i in range(0,len(string)-2):
if string[i:i+len(sub_string)] == sub_string:
c+=1
return c
if __name__ == '__main__':
string = input().strip()
sub_string = input().strip()
count = count_substring(string, sub_string)
print(count)
I runned in the same problem and did this:
hw = 'Hello oh World!'
list_hw = list(hw)
o_in_hw = []
while True:
o = hw.find('o')
if o != -1:
o_in_hw.append(o)
list_hw[o] = ' '
hw = ''.join(list_hw)
else:
print(o_in_hw)
break
Im pretty new at coding so you can probably simplify it (and if planned to used continuously of course make it a function).
All and all it works as intended for what i was doing.
Edit: Please consider this is for single characters only, and it will change your variable, so you have to create a copy of the string in a new variable to save it, i didnt put it in the code cause its easy and its only to show how i made it work.
By slicing we find all the combinations possible and append them in a list and find the number of times it occurs using count function
s=input()
n=len(s)
l=[]
f=input()
print(s[0])
for i in range(0,n):
for j in range(1,n+1):
l.append(s[i:j])
if f in l:
print(l.count(f))
To find all the occurence of a character in a give string and return as a dictionary
eg: hello
result :
{'h':1, 'e':1, 'l':2, 'o':1}
def count(string):
result = {}
if(string):
for i in string:
result[i] = string.count(i)
return result
return {}
or else you do like this
from collections import Counter
def count(string):
return Counter(string)
I have a string like
"xdtwkeltjwlkejt7wthwk89lk"
how can I get the index of the first digit in the string?
Use re.search():
>>> import re
>>> s1 = "thishasadigit4here"
>>> m = re.search(r"\d", s1)
>>> if m:
... print("Digit found at position", m.start())
... else:
... print("No digit in that string")
...
Digit found at position 13
Here is a better and more flexible way, regex is overkill here.
s = 'xdtwkeltjwlkejt7wthwk89lk'
for i, c in enumerate(s):
if c.isdigit():
print(i)
break
output:
15
To get all digits and their positions, a simple expression will do
>>> [(i, c) for i, c in enumerate('xdtwkeltjwlkejt7wthwk89lk') if c.isdigit()]
[(15, '7'), (21, '8'), (22, '9')]
Or you can create a dict of digit and its last position
>>> {c: i for i, c in enumerate('xdtwkeltjwlkejt7wthwk89lk') if c.isdigit()}
{'9': 22, '8': 21, '7': 15}
Thought I'd toss my method on the pile. I'll do just about anything to avoid regex.
sequence = 'xdtwkeltjwlkejt7wthwk89lk'
i = [x.isdigit() for x in sequence].index(True)
To explain what's going on here:
[x.isdigit() for x in sequence] is going to translate the string into an array of booleans representing whether each character is a digit or not
[...].index(True) returns the first index value that True is found in.
Seems like a good job for a parser:
>>> from simpleparse.parser import Parser
>>> s = 'xdtwkeltjwlkejt7wthwk89lk'
>>> grammar = """
... integer := [0-9]+
... <alpha> := -integer+
... all := (integer/alpha)+
... """
>>> parser = Parser(grammar, 'all')
>>> parser.parse(s)
(1, [('integer', 15, 16, None), ('integer', 21, 23, None)], 25)
>>> [ int(s[x[1]:x[2]]) for x in parser.parse(s)[1] ]
[7, 89]
import re
first_digit = re.search('\d', 'xdtwkeltjwlkejt7wthwk89lk')
if first_digit:
print(first_digit.start())
To get all indexes do:
idxs = [i for i in range(0, len(string)) if string[i].isdigit()]
Then to get the first index do:
if len(idxs):
print(idxs[0])
else:
print('No digits exist')
As the other solutions say, to find the index of the first digit in the string we can use regular expressions:
>>> s = 'xdtwkeltjwlkejt7wthwk89lk'
>>> match = re.search(r'\d', s)
>>> print match.start() if match else 'No digits found'
15
>>> s[15] # To show correctness
'7'
While simple, a regular expression match is going to be overkill for super-long strings. A more efficient way is to iterate through the string like this:
>>> for i, c in enumerate(s):
... if c.isdigit():
... print i
... break
...
15
In case we wanted to extend the question to finding the first integer (not digit) and what it was:
>>> s = 'xdtwkeltjwlkejt711wthwk89lk'
>>> for i, c in enumerate(s):
... if c.isdigit():
... start = i
... while i < len(s) and s[i].isdigit():
... i += 1
... print 'Integer %d found at position %d' % (int(s[start:i]), start)
... break
...
Integer 711 found at position 15
In Python 3.8+ you can use re.search to look for the first \d (for digit) character class like this:
import re
my_string = "xdtwkeltjwlkejt7wthwk89lk"
if first_digit := re.search(r"\d", my_string):
print(first_digit.start())
I'm sure there are multiple solutions, but using regular expressions you can do this:
>>> import re
>>> match = re.search("\d", "xdtwkeltjwlkejt7wthwk89lk")
>>> match.start(0)
15
Here is another regex-less way, more in a functional style. This one finds the position of the first occurrence of each digit that exists in the string, then chooses the lowest. A regex is probably going to be more efficient, especially for longer strings (this makes at least 10 full passes through the string and up to 20).
haystack = "xdtwkeltjwlkejt7wthwk89lk"
digits = "012345689"
found = [haystack.index(dig) for dig in digits if dig in haystack]
firstdig = min(found) if found else None
you can use regular expression
import re
y = "xdtwkeltjwlkejt7wthwk89lk"
s = re.search("\d",y).start()
def first_digit_index(iterable):
try:
return next(i for i, d in enumerate(iterable) if d.isdigit())
except StopIteration:
return -1
This does not use regex and will stop iterating as soon as the first digit is found.
import re
result = " Total files:................... 90"
match = re.match(r".*[^\d](\d+)$", result)
if match:
print(match.group(1))
will output
90
instr = 'nkfnkjbvhbef0njhb h2konoon8ll'
numidx = next((i for i, s in enumerate(instr) if s.isdigit()), None)
print(numidx)
Output:
12
numidx will be the index of the first occurrence of a digit in instr. If there are no digits in instr, numidx will be None.
I didn't see this solution here, and thought it should be.