How to Print a List of Elements Separated by Commas

Printing a list separated with commas, without a trailing comma

Pass sep="," as an argument to print()

You are nearly there with the print statement.

There is no need for a loop, print has a sep parameter as well as end.

>>> print(*range(5), sep=", ")
0, 1, 2, 3, 4

A little explanation

The print builtin takes any number of items as arguments to be printed. Any non-keyword arguments will be printed, separated by sep. The default value for sep is a single space.

>>> print("hello", "world")
hello world

Changing sep has the expected result.

>>> print("hello", "world", sep=" cruel ")
hello cruel world

Each argument is stringified as with str(). Passing an iterable to the print statement will stringify the iterable as one argument.

>>> print(["hello", "world"], sep=" cruel ")
['hello', 'world']

However, if you put the asterisk in front of your iterable this decomposes it into separate arguments and allows for the intended use of sep.

>>> print(*["hello", "world"], sep=" cruel ")
hello cruel world

>>> print(*range(5), sep="---")
0---1---2---3---4

Using join as an alternative

The alternative approach for joining an iterable into a string with a given separator is to use the join method of a separator string.

>>>print(" cruel ".join(["hello", "world"]))
hello cruel world

This is slightly clumsier because it requires non-string elements to be explicitly converted to strings.

>>>print(",".join([str(i) for i in range(5)]))
0,1,2,3,4

Brute force - non-pythonic

The approach you suggest is one where a loop is used to concatenate a string adding commas along the way. Of course this produces the correct result but its much harder work.

>>>iterable = range(5)
>>>result = ""
>>>for item, i in enumerate(iterable):
>>>    result = result + str(item)
>>>    if i > len(iterable) - 1:
>>>        result = result + ","
>>>print(result)
0,1,2,3,4

How can I print a list of elements separated by commas?

Use an infix_iterator:

// infix_iterator.h 
// 
// Lifted from Jerry Coffin's 's prefix_ostream_iterator 
#if !defined(INFIX_ITERATOR_H_) 
#define  INFIX_ITERATOR_H_ 
#include <ostream> 
#include <iterator> 
template <class T, 
          class charT=char, 
          class traits=std::char_traits<charT> > 
class infix_ostream_iterator : 
    public std::iterator<std::output_iterator_tag,void,void,void,void> 
{ 
    std::basic_ostream<charT,traits> *os; 
    charT const* delimiter; 
    bool first_elem; 
public: 
    typedef charT char_type; 
    typedef traits traits_type; 
    typedef std::basic_ostream<charT,traits> ostream_type; 
    infix_ostream_iterator(ostream_type& s) 
        : os(&s),delimiter(0), first_elem(true) 
    {} 
    infix_ostream_iterator(ostream_type& s, charT const *d) 
        : os(&s),delimiter(d), first_elem(true) 
    {} 
    infix_ostream_iterator<T,charT,traits>& operator=(T const &item) 
    { 
        // Here's the only real change from ostream_iterator: 
        // Normally, the '*os << item;' would come before the 'if'. 
        if (!first_elem && delimiter != 0) 
            *os << delimiter; 
        *os << item; 
        first_elem = false; 
        return *this; 
    } 
    infix_ostream_iterator<T,charT,traits> &operator*() { 
        return *this; 
    } 
    infix_ostream_iterator<T,charT,traits> &operator++() { 
        return *this; 
    } 
    infix_ostream_iterator<T,charT,traits> &operator++(int) { 
        return *this; 
    } 
};     
#endif 

Usage would be something like:

#include "infix_iterator.h"

// ...
std::copy(keywords.begin(), keywords.end(), infix_iterator(out, ","));

Haskell: how to print each element of list separated with comma

You can use intercalate. It'll insert the comma between each element of the list and concatenate the resulting list of strings to turn it into a single string.

import Data.List

toCommaSeparatedString :: [String] -> String
toCommaSeparatedString = intercalate ","

ghci> toCommaSeparatedString ["1","2","3"]
"1,2,3"

print list and separate values by a comma comma and end with a dot

I would approach this by maintaining some additional state which keeps track of whether or not it is the first open locker which needs to be reported. And then, just print period outside the loop, only once.

printf("Open lockers: ");
int first = 1;

for (int i=0; i < sizeof(lockers); i++) {
    if (lockers[i] == true) {
        if (first == 0) {
            printf(", ");
        }
        else {
            first = 0;
        }

        printf("%d", i + 1);
    }
}
printf(".");

Demo

Note: In the demo I replaced your bool lockers array with an int array. But the rest of the logic remains the same.

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