I need to generate a string from a float which is always the length of 5. For example:
input_number: float = 2.22
output_str = "00222"
The float never larger then 999.xx and can have an arbitrary number of decimal places. I came up with the following code, but I doubt whether what I have in mind can't be done in a more pythonic way.
My solution:
input_number = 343.2423423
input_rounded = round(input_number, 2)
input_str = str(input_rounded)
input_str = input_str.replace(".","")
input_int = int(input_str)
output_str = f"{input_int:05d}"
More examples:
343.2423423 -> "34324"
23.3434343 -> "02334"
CodePudding user response:
Does this match your use cases:
for input_number in (343.2423423, 23.3434343, 0.34, .1):
num, decimal = str(input_number).split('.')
print(f"{num.zfill(3)}{decimal[:2].ljust(2, '0')}")
Out:
34324
02334
00034
00010
CodePudding user response:
This is the simplest way I can think of. Format it into a string with two decimal places shown and six total characters (left-filling with zeroes), then remove the decimal point.
def format_number(number):
return f'{number:06.2f}'.replace('.', '')
The format specification mini-language can seem pretty arcane, but in this case it's just:
0
: the character to fill with6
: the total number of characters to produce (including the decimal).2
: the number of digits to show after the decimalf
: fixed-point float notation
CodePudding user response:
Here is one way to do it:
for input_number in (343.2423423, 23.3434343, 0.34, .1, 23.45, 2.32):
num, decimal = str(input_number).split('.')
formatted_num = "{:03d}".format(int(num))
formatted_decimal = "{:.2f}".format(float("0." decimal))[2:]
print(formatted_num formatted_decimal)