Welcome to the Lounge

lopatir wrote:

but how do you misspell "i"

class StringFormat(object):

    @staticmethod
    def FloatToStringFormat(num: float, decimalPlaces: int) -> str:
        return "{:.{}f}".format(num, decimalPlaces)

resultAsString = StringFormat.FloatToStringFormat(20.123, "2")

Mike Winiberg wrote:

has almost nothing to do with the language you write it in, and everything to do with domain knowledge and the skill of the programmer(s).

Dictionary<string, IEnumerable<IEnumerable<SomeValue>>> valuesPerLineFromACsvFileByKey = new Dictionary<string, IEnumerable<IEnumerable<SomeValue>>>();

var valuesPerLineFromACsvFileByKey = new Dictionary<string, IEnumerable<IEnumerable<SomeValue>>>();

if x is a:
   do //...

if x is a:
	do //...

So maybe we should shut up and accept that Python Is, After All, Doing The Right Thing?

If your attitude is that "Noone should keep on critizising my language of choice!", then maybe you should search for another language.

you think assigning semantics to use of whitespace, you should take a look at the Whitespace language

Nothing related to indentation has any semantics in the C class of languages, so pretending that they are similar is a complete misunderstanding.

A week or two ago, someone posted an April 1st proposal from Bjarne Strostrup (I am assuming that it really was written by him), a proposal about extending the operatior definitinions of C++, to allow operators to be defined for whitespace. He goes on to propose operator definitions for lack of whitespace. Sometimes, I suspect that Python addicts never noticed the publication date.

# Python program to check if the input number is prime or not

num = 15

# take input from the user
# num = int(input("Enter a number: "))

# prime numbers are greater than 1

if num > 1:
    for i in range(2,num):
        if (num % i) == 0:
            print(num,"is not a prime number")	   
            print(i,"times",num//i,"is",num)
            break
    else:
            #print(num % i)
            print(num,"is a prime number")
            
# if input number is less than
# or equal to 1, it is not prime
else:
   print(num,"is not a prime number")