How to Remove Zeros After Decimal from String Remove All Zero After Dot

how to remove zeros after decimal from string remove all zero after dot

If want convert integers and floats numbers to strings with no trailing 0 use this with map or apply:

df = pd.DataFrame({'col1':[1.00, 1, 0.5, 1.50]})

df['new'] = df['col1'].map('{0:g}'.format)
#alternative solution
#df['new'] = df['col1'].apply('{0:g}'.format)
print (df)
col1 new
0 1.0 1
1 1.0 1
2 0.5 0.5
3 1.5 1.5

print (df['new'].apply(type))
0 <class 'str'>
1 <class 'str'>
2 <class 'str'>
3 <class 'str'>
Name: new, dtype: object

Remove useless zero digits from decimals in PHP

$num + 0 does the trick.

echo 125.00 + 0; // 125
echo '125.00' + 0; // 125
echo 966.70 + 0; // 966.7

Internally, this is equivalent to casting to float with (float)$num or floatval($num) but I find it simpler.

Regex for remove unnecessary zeros after decimal point

You don't need a regex for this. Just cast your string to double:

$arr = array('21250000.022000', '20.00', '200', '20.50');

foreach ($arr as $n)
echo (double) $n. "\n";

Output:

21250000.022
20
200
20.5

Code Demo


Update: If you're looking for a regex solution then use:

Search regex:

(?:(\.\d*?[1-9]+)|\.)0*$

replacement:

$1

RegEx Demo

how to remove zeros after decmal from string remove all zero after dot

convert it into integer type..............................................

oR

Split the string, use decimal point(.) as a spliter character.

OR

String val=Html.fromHtml(ProductList.Category_price.get(position));
String val1=val.substring(0,val.indexOf("."));
holder.txtText3.setText("Price: "+val1);

Remove insignificant trailing zeros from a number?

If you convert it to a string it will not display any trailing zeros, which aren't stored in the variable in the first place since it was created as a Number, not a String.

var n = 1.245000
var noZeroes = n.toString() // "1.245"

How to remove trailing zeros using Dart

I made regular expression pattern for that feature.

double num = 12.50; // 12.5
double num2 = 12.0; // 12
double num3 = 1000; // 1000

RegExp regex = RegExp(r'([.]*0)(?!.*\d)');

String s = num.toString().replaceAll(regex, '');

Remove extra zeros from a string

It occurs to me you can do a single find and replace to accomplish

everything.

With this, you match the entire valid number at a time, enabling you to

fix multiple numbers in a single string, if done globally.

Also works the same if your string contains a single number.

Find (?:(-)(?![0.]+(?![\d.]))|-)?\d*?([1-9]\d*|0)(?:(?:(\.\d*[1-9])|\.)\d*)?(?![\d.])

Replace $1$2$3

JS demo: https://regex101.com/r/H44t6z/1

Readable / Info version

 # Add behind boundary check here
# -----------------
(?:
( - ) # (1), Preserve sign -
(?! # Only if not a zero value ahead
[0.]+
(?! [\d.] )
)
| # or
- # Match sign, but dump it
)?
\d*? # Dump leading 0's
( # (2 start), Preserve whole number
[1-9] # First non-0 number
\d* # Any number
| # or
0 # Just last 0 before decimal
) # (2 end)
(?: # Optional fraction part
(?: # -------------
( # (3 start), Preserve decimal and fraction
\. # Decimal
\d* # Any number
[1-9] # Last non-0 number
) # (3 end)
| # or
\. # Match decimal, but dump it
) # -------------
\d* # Dump trailing 0's
)?
(?! [\d.] ) # No digits or dot ahead

Remove trailing zero in C++

This is one thing that IMHO is overly complicated in C++. Anyway, you need to specify the desired format by setting properties on the output stream. For convenience a number of manipulators are defined.

In this case, you need to set fixed representation and set precision to 2 to obtain the rounding to 2 decimals after the point using the corresponding manipulators, see below (notice that setprecisioncauses rounding to the desired precision). The tricky part is then to remove trailing zeroes. As far as I know C++ does not support this out of the box, so you have to do some string manipulation.

To be able to do this, we will first "print" the value to a string and then manipulate that string before printing it:

#include <iostream>
#include <iomanip>

int main()
{
double value = 12.498;
// Print value to a string
std::stringstream ss;
ss << std::fixed << std::setprecision(2) << value;
std::string str = ss.str();
// Ensure that there is a decimal point somewhere (there should be)
if(str.find('.') != std::string::npos)
{
// Remove trailing zeroes
str = str.substr(0, str.find_last_not_of('0')+1);
// If the decimal point is now the last character, remove that as well
if(str.find('.') == str.size()-1)
{
str = str.substr(0, str.size()-1);
}
}
std::cout << str << std::endl;
}

R str_replace / remove access zeros in string after the decimal seperator, (zero following dot[1-9]) gesub, regex, regular expression

You could try to find:

\b(\d+)(?:\.0+|(\.\d+?))0*\b

And replace with \1\2. See an online demo



  • \b - Word-boundary;
  • (\d+) - Capture trailing digits upto;
  • (?: - Open a non-capture group;
    • \.0+ - 1+ zero's;
    • | - Or;
    • (\.\d+?) - A nested 2nd capture group to match a dot followed by a digit and 0+ (lazy) digits;
    • ) - Close non-capture group;
  • 0* - 0+ (greedy) digits;
  • \b - Trailing word-boundary.


library(stringr)
v <- c("bla 500.00", "bla 1.20", "bla 1.10", "bla 2.34", "bla 2.340003", "bla 1.032", "bla 1.10 bla 2.00")
v <- str_replace_all(v, "\\b(\\d+)(?:\\.0+|(\\.\\d+?))0*\\b", "\\1\\2")
v

Prints: "bla 500", "bla 1.2", "bla 1.1", "bla 2.34", "bla 2.340003", "bla 1.032", "bla 1.1 bla 2"



Related Topics



Leave a reply



Submit