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Python is adding point zero at end post number to string

I am unable to get why point zero is adding at the end of the string post converting from int to string.
The below is the code which I am trying to convert:
df['attendees_num'].astype(str)
here attendees_num value is 20 but its giving output 20.0. can someone help me how do i convert it into 20 as string without .0 at end.?

This should do it:
df['attendees_num'].astype(int).astype(str)

How can I pack a string in struct.pack

I am trying to pack a string using struct.pack.
I am able to see complete value if I use integer type but when I want to use string I only see one character.
struct.pack("<1L",0xabcdabcd)
'\xab\xcd\ab\cd'
struct.pack("<1s","overflow")
'o' prints just s. I wanted it to print full string: overflow.

In the format string ("<1s") you're passing to struct.pack, the 1 denotes the maximum number of characters that field can store. (See the paragraph beginning "For the 's'..." in the struct documentation.) Since you're passing 1, it will only store the first character. You'll need to choose a length that will fit any string you want to store in the struct, and specify that. For example, to store the string "overflow" (8 characters) you could use "<8s":
>>> struct.pack("<8s", "overflow")
'overflow'

Why Simple String's Character has So many array Dimensions?

I am currently working on a python's String and List.
When I assign string in variable str="string" and try to access it first character by str[0] it works perfectly and give "s".
But, when I try to find character str[0][0][0][0][0][0] it again gives "s". But when I give str[0][1] it gives an error:
IndexError: string index out of range
Its Correct. My Question is Why Simple String Character has So many array Dimensions? and it did not given any error and print 0 character of string when str[0][0][0][0][0][0]. What is Data Structure of String?
My Code is
str="string"
print((str[0][0][0][0][0][0][0][0])) # Working, but my Question is Why Working
print((str[1][0][0][0][0])) # Working
print((str[2][0][0][0][0])) # Working
print((str[3][0][0][0][0])) # Working
list=["0","p",0]
print(list[0][0][0]) # Working
My Output is:
s
t
r
i
0

Why shouldn't it work?
Indexing a string returns a one element string which is again indexable and returns the same value:
>>> 's'[0]
's'
since it consists of one element, you can continue indexing the zero-element [0] as much as you want.
This is explained in the standard type hierarchy section of the Python Reference manual:
Strings
A string is a sequence of values that represent Unicode code points. All the code points in the range U+0000 - U+10FFFF can be represented in a string. Python doesn’t have a char type; instead, every code point in the string is represented as a string object with length 1.
(Emphasis mine)
Side-note: Don't use names such as str, you mask the built-in str.

In Python a string is a sequence of characters, but characters are 1-char strings.
So if you access 'foobar'[0], you obtain 'f'. Since f is however a string, we can access the first character of that string. Since 'f'[0] is 'f'. As a result if you access a strings s with s[i][0][0][0], you thus keep accessing the first character of the string s.
If you write s[i][1] however, this will error, since s[i] is a one-character string, and thus you can not obtain the second character, since there is no such character.
The string itself is not multidimensional, you simply obtain a new string and call the index of that new string. You can add as many [0]s as you like.

The problem is not in Python, it is due to the fact that you assume there is a char type in Python (based on the title of this question).
A string in Python is an array of essentially single element strings. s[0] simply returns the string 's', not a character. s[0]...[0] can be thought of as an infinite recursion that keeps getting the same single element string, infinitely many times.
You can go as deep as you want: (in this case, in order to do it more than 997 times you will need to modify Python`s default allowed recursion depth)
def string_dive(s, count=0):
if count < 997:
count += 1
return string_dive(s[0], count)
else:
return s
print(string_dive('string'))
# 's'

ValueError when using long() in Python

So when I run:
value = long("00000000000000020000000000000002", 16)
I get :
ValueError: Value out of range: 36893488147419103234
I think it's because long can't take such a big hex number, but I'm not sure.
In reality I'm iterating through a file with a large amount of very big hex numbers, but this is just an example of one of the hex numbers I'm trying to parse.
I've tried using lstrip() to remove some of the 0's but it made no difference to the error.
What am I doing wrong?

The error was being caused by the variable I was trying to assign the value to, not the actual long() function.

Import string that looks like a list "[0448521958, +61439800915]" from JSON into Python and make it an actual list?

I am extracting a string out of a JSON document using python that is being sent by an app in development. This question is similar to some other questions, but I'm having trouble just using x = ast.literal_eval('[0448521958, +61439800915]') due to the plus sign.
I'm trying to get each phone number as a string in a python list x, but I'm just not sure how to do it. I'm getting this error:
raise ValueError('malformed string')
ValueError: malformed string

your problem is not just the +
the first number starts with 0 which is an octal number ... it only supports 0-7 ... but the number ends with 8 (and also has other numbers bigger than 8)
but it turns out your problems dont stop there
you can use regex to fix the plus
fixed_string = re.sub('\+(\d+)','\\1','[0445521757, +61439800915]')
ast.literal_eval(fixed_string)
I dont know what you can do about the octal number problem however

I think the problem is that ast.literal_eval is trying to interpret the phone numbers as numbers instead of strings. Try this:
str = '[0448521958, +61439800915]'
str.strip('[]').split(', ')
Result:
['0448521958', '+61439800915']

Technically that string isn't valid JSON. If you want to ignore the +, you could strip it out of the file or string before you evaluate it. If you want to preserve it, you'll have to enclose the value with quotes.