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How to put a str into another str, but on a specific digit in the other str? [duplicate]

This question already has answers here:
Insert some string into given string at given index [duplicate]
(9 answers)
Closed last year.
(Example:)
a = "-"
b = "testtest"
I want to put the a = "-" after 4 digits of the bstring (so between the first and second "test").
So it should be always after the first 4 digits (if the b-str is longer, it shouldn't just go at the first 4th digit.

Apply slicing in a string
index = 4
a = '-'
b = 'testtest'
c = b[:index] + a + b[index:]

a = "-"
b = "testtest"
c = b[:4]+a+b[4:]
print (c)

how to count the letters in a string? [duplicate]

This question already has answers here:
How to count digits, letters, spaces for a string in Python?
(12 answers)
Closed 5 years ago.
how do I code a program that would produce this below?
As an example, the following code fragment:
print (count_letters("ab1c2d345"))
should produce the output:
4

You can try this:
def count_letters(s):
return len([i for i in s if i.isalpha()])
print(count_letters('ab1c2d345'))
Output:
4
Or, you can use regex for a cleaner solution:
import re
def count_letters(s):
return len(re.findall('[a-zA-Z]', s))

You can do it using simple loop/if statement.
def count_letters(str_ltr):
count_ltr = 0
for ch in str_ltr:
if ch.isalpha():
count_ltr += 1
return count_ltr
print(count_letters("ab1c2d345"))
Output: 4

How to count the number of a specific character at the end of a string ignoring duplicates? [duplicate]

This question already has answers here:
Pythonic way to count the number of trailing zeros in base 2
(5 answers)
Closed 5 years ago.
I have a series of strings like:
my_text = "one? two three??"
I want to count only the number of ? at the end of the string. The above should return 2 (rather than 3).
What I've tried so far:
my_text.count("?") # returns 3

There's not a built-in method for it. But something simple like this should do the trick:
>>> len(my_text) - len(my_text.rstrip('?'))
2

You could also use a regexp to count the number of trailing question marks :
import re
def count_trailing_question_marks(text):
last_question_marks = re.compile("\?*$")
return len(last_question_marks.search(text).group(0))
print count_trailing_question_marks("one? two three??")
# 2
print count_trailing_question_marks("one? two three")
# 0

Not so clean but simple way:
my_text = "one? two three??"
total = 0
question_mark = '?'
i = 0
for c in my_text:
i -= 1
if my_text[i] == question_mark:
total += 1
else:
break

One-liner using my favourite itertools:
First reverse the string, then continue iterating (taking the values) while our condition is satisfied (value == '?'). This returns an iterable which we exhaust into a list and finally take its length.
len(list(itertools.takewhile(lambda x:x=='?',reversed(my_text))))

Why is it printing an # instead of \100? [duplicate]

This question already has answers here:
Why do numbers in a string become "x0n" when a backslash precedes them?
(2 answers)
the reason: python string assignments accidentally change '\b' into '\x08' and '\a' into '\x07', why Python did this?
(2 answers)
Closed 7 years ago.
I have the following program:
x = 0
while x <= 10:
print(x, '\10')
x = x + 1
It then prints:
0 #
1 #
2 #
3 #
4 #
5 #
6 #
7 #
8 #
9 #
Instead of:
1 \ 10, 2 \ 10 And so on...
Why is the program doing this?

You're escaping the 10 with a \ symbol and python is interpreting \10 as a code for the # symbol instead. You can fix this by either placing an r character as a prefix for the string or by escaping the backslash with another one.
Fix:
r'\10' #Raw string
Or:
'\\10'

Count of a string in a list [duplicate]

This question already has answers here:
How do I count the occurrences of a list item?
(30 answers)
Closed 9 years ago.
I need help:
I want to write function that: takes a list as a input and return count of specific string.
If function input is x=[a,a,b,c] function need return 2 (a is two times in this list).

>>> def F(seq, string):
return seq.count(string)
>>> F(['a','a','b','c'], 'a')
2
Equivalent to:
def F(seq, string):
n = 0
for x in seq:
if x == string:
n += 1
return n