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Spliting string after certain amount of characters

I have a lengthy string and want to split it after a certain number of characters. I already have done this:
if len(song.lyrics) > 2048:
string1 = string[:2048]
string2 = string[2049:]
The problem with this is that sometimes it breaks in the middle of text and I don't want to. Is there a way to get the last linebreak before the character limit is reached and break it there?
Thanks

Does this give you the result you're looking for? If not, could you please provide an example string with expected output?
import re
CHARACTER_LIMIT = 2048
for m in re.finditer(r'.{,%s}(?:\n|$)' % CHARACTER_LIMIT, string, re.DOTALL):
print(m.group(0))

Find the index of newline character just-left-of your length limit then use it to split.
if len(song.lyrics) > 2048:
index = string[:2048].rfind('\n')
string1 = string[:index]
string2 = string[index+1:]
Example:
>>> s = 'aaaaaaa\nbbbbbbbbbbbbbbbb\nccccccc\ndddddddddddddddd'
>>> limit = 31 # ↑
>>> index = s[:limit].rfind('\n')
>>> index
24
>>> s1,s2 = s[:index],s[index+1:]
>>> s1
'aaaaaaa\nbbbbbbbbbbbbbbbb'
>>> s2
'ccccccc\ndddddddddddddddd'
>>>

How to count the number of characters at the start of a string?

How can I count the number of characters at the start/end of a string in Python?
For example, if the string is
'ffffhuffh'
How would I count the number of fs at the start of the string? The above string with a f should output 4.
str.count is not useful to me as a character could be in the middle of the string.

A short and simple way will be to use the str.lstrip method, and count the difference of length.
s = 'ffffhuffh'
print(len(s)-len(s.lstrip('f')))
# output: 4
str.lstrip([chars]):
Return a copy of the string with leading characters removed. The chars
argument is a string specifying the set of characters to be removed.

Try this, using itertools.takewhile():
import itertools as it
s = 'ffffhuffh'
sum(1 for _ in it.takewhile(lambda c: c == 'f', s))
=> 4
Similarly, for counting the characters at the end:
s = 'huffhffff'
sum(1 for _ in it.takewhile(lambda c: c == 'f', reversed(s)))
=> 4

You may use regular expression with re.match to find the occurrence of any character at the start of the string as:
>>> import re
>>> my_str = 'ffffhuffh'
>>> my_char = 'f'
>>> len(re.match('{}*'.format(my_char), my_str).group())
4

Building on Oscar Lopez's answer, I want to handle the case you mention of the end of the string: use reversed()
import itertools as it
my_string = 'ffffhuffh'
len(list(it.takewhile(lambda c: c == my_string[-1], reversed(my_string))))
=> 1

You can create a function and iterate through your string and return the count of the desired char in the input string's beginning or end like this example:
# start = True: Count the chars in the beginning of the string
# start = False: Count the chars in the end of the string
def count_char(string= '', char='', start=True):
count = 0
if not start:
string = string[::-1]
for k in string:
if k is char:
count += 1
else:
break
return count
a = 'ffffhuffh'
print(count_char(a, 'f'))
b = a[::-1]
print(count_char(b, 'f', start=False))
Output:
4
4

You may also use itertools.groupby to find the count of the occurrence of the first element at the start of the string as:
from itertools import groupby
def get_first_char_count(my_str):
return len([list(j) for _, j in groupby(my_str)][0])
Sample run:
>>> get_first_char_count('ffffhuffh')
4
>>> get_first_char_count('aywsnsb')
1

re.sub select first letter with repeat( (^(\w)\2*) ), len count frequency.
len(re.sub(r'((^\w)\2*).*',r'\1',my_string))

Convert an integer into a string of its ascii values

Given a number number such that its digits are grouped into parts of length n (default value of n is 3) where each group represents some ascii value, I want to convert number into a string of those ascii characters. For example:
n number Output
==================================
3 70 F
3 65066066065 ABBA
4 65006600660065 ABBA
Note that there is no leading 0 in number, so the first ascii value will not necessarily be represented with n digits.
My current code looks like this:
def number_to_string(number, n=3):
number = str(number)
segment = []
while number:
segment.append(number[:n])
number = number[n:]
return str(''.join('{:0>{}}'.format(chr(segment), n) for segment in number))
Expected outputs:
number_to_string(70)
'F'
number_to_string(65066066065)
'ABBA'
number_to_string(65006600660065, n=4)
'ABBA'
My current code however returns an empty string. For example, instead of 'F' it returns ' '. Any reason why this is? Thank you!
P.S.:
I'm wanting to reverse the process of this question, i.e. turn an integer into a string based on the ascii values of each character (number) in the string. But reading that question is not a requirement to answer this one.

Try this:
import re
def number_to_string(num, n=3):
num_str = str(num)
if len(num_str) < n:
num_str = '0' * (n-len(num_str)) + num_str
elif len(num_str) % n != 0:
num_str = '0'*(n-len(num_str)%n) + num_str
print(num_str)
chars = re.findall('.'*n, num_str)
l = [chr(int(i)) for i in chars]
return ''.join(l)
First pad the given number (converted into string) with required number of zeros, so that it can be evenly split into equal number of characters each. Then using re split the string into segments of size n. Finally convert each chunk into character using chr, and then join them using join.

def numToStr(inp):
"""Take a number and make a sequence of bytes in a string"""
out=""
while inp!=0:
out=chr(inp & 255)+out
inp=inp>>8
print "num2string:", out
return out
does this help?

Is this what you want?
def num_to_string(num, leng):
string = ""
for i in range(0,len(str(num)),leng):
n = str(num)[i:i+2]
string += chr(int(n))
print string
Output:
>>> ================================ RESTART ================================
>>>
>>> num_to_string(650065006600,4)
AAB
>>> num_to_string(650650660,3)
AAB
>>> num_to_string(656566,2)
AAB
>>>

You can just append \x to number as this prints 'p':
print '\x70'

Check, using a list, if a string contains a certain letter or number

The question I'm answering requires you to validate a Car Reg Plate. It must look for a sequence of two letters, then three numbers, then three letters, otherwise return a message like "not a valid postcode"
I need to know how to check if a string contains a certain letter, or number, by comparing it with a list.
So far, I've got this:
# Task 2
import random
import re
def regNumber():
# Generate a Car Reg Number
letters = ["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"]
numbers = ["1","2","3","4","5","6","7","8","9","0"]
letter1 = random.choice(letters)
letter2 = random.choice(letters)
number1 = random.choice(numbers)
number2 = random.choice(numbers)
letter3 = random.choice(letters)
letter4 = random.choice(letters)
letter5 = random.choice(letters)
licensePlate = (letter1 + letter2 + number1 + number2 + letter3 + letter4 + letter5)
return licensePlate, letters, numbers
carReg, letters, numbers = regNumber()
print(carReg)
if letters not in carReg: print("Success")
However, I get this error:
TypeError: 'in <string>' requires string as left operand, not list
Any help appreciated.

You need to be checking for characters in your string with this method, it will not simply iterate over your list for you.
Try using something like this instead, to check every character in your list of strings:
if any(letter in carReg for letter in letters):
This will cut out on the first True, which is I think what you're looking for.
Note: If using any like this is unfamiliar territory for you, you can also always just iterate over every string within your list of strings to check for those given characters.
Update: If you're attempting to match a given format of letters and numbers, it would make much more sense (IMHO) for you to familiarize yourself with Python's regex methods to pattern match to a valid license plate than attempt to use loops to validate one. I won't write the regex for your particular case, but to give you an idea, the following would allow you to match 3 letters followed by 1-4 digits (valid license plate where I live)
match_plate = re.compile(r"^[A-Z]{3}\d{1,4}$",re.I)
If you really must use a list to check, you will have to use a series of conditional statements to split the license plate into parts over which you can validate with iterations.

The error is telling you the exact issue in this case,
letters is a list being returend from regNumber but in requires a string on the leftside
like 'ASD111' in carReg

change
if letters not in carReg: print("Success")
to
for l in letters:
if l not in carReg:
print("Success")
in your code you are having a list of strings and, that is why I have changed your if condition to a for loop so that each element of the list is checked for occurance in carReg string.
alternatively, i think you should be using a flag to solve your probem. Like so:
flag = 0
for l in letters:
if l in carReg:
flag = 1
break
if flag == 0:
print("Success")

Another way in which you could generate a certain number of letters rather than having to use so many variables would be to use just two variables that would allow the generation of, for the first one, 2 letters and for the second 3 letters.
An example of how I would implement this would be:
def randomLetters1(y):
return ''.join(random.choice(string.ascii_uppercase) for x in range(y))
firstLetters = (randomLetters1(2))
secondLetters = (randomLetters1(3))
I know this because I have had to do this exact same task.

You could do it without regular expressions:
Define the pattern you want using str methods in a list
pattern = [str.isalpha, str.isalpha,
str.isdigit, str.isdigit, str.isdigit,
str.isalpha, str.isalpha, str.isalpha]
Use that pattern to check a string.
def is_foo(pattern, s):
'''Return True if s matches pattern
s is a string
pattern is a list of functions that return a boolean
len(s) == len(pattern)
each function in pattern is applied to the corresponding character in s
'''
# assert len(s) == len(pattern)
return all(f(c) for f, c in zip(pattern, s))
Usage
if is_foo(pattern, 'aa111aaa'):
print 'yes!!!'
if not is_foo(pattern, '11aa111'):
print ':('

clear and comprehensible way to calculate the string [12:3]

I new on python.
I have this string "[12:3]" and i what to calculate the difference between these two numbers.
Ex: 12 - 3 = 9
Of course I can do something (not very clear) like this:
num1 = []
num2 = []
s = '[12:3]'
dot = 0;
#find the ':' sign
for i in range(len(s)):
if s[i] == ':' :
dot = i
#left side
for i in range(dot):
num1.append(s[i])
#right side
for i in range(len(s) - dot-1):
num2.append(s[i+dot+1])
return str(int("".join(num1))-int("".join(num2))+1)
But i'm sure the is a more clear and comprehensible way.
Thanks!

You could use regex to pick the numbers out of your string:
import re
s = '[12:3]'
numbers = [int(x) for x in re.findall(r'\d+',s)]
return numbers[0]-numbers[1]

Or, without re
numbers = [int(x) for x in s.strip('[]').split(':')]
print numbers[0] - numbers[1]
prints
9

You should use regular expressions.
>>> import re
>>> match = re.match(r'\[(\d+):(\d+)\]', '[12:3]')
>>> match.groups()
('12', '3')
>>> a = int(match.groups()[0])
>>> b = int(match.groups()[1])
>>> a - b
9
The regular expression there says "match starting at the beginning of the string, find [, then any number of digits \d+ (and store them), then a :, then any number of digits \d+ (and store them), and finally ]". We then extract the stored digits using .groups() and do arithmetic on them.