How to "properly" print a list?

Question

So I have a list:

['x', 3, 'b']

And I want the output to be:

[x, 3, b]

How can I do this in python?

If I do str(['x', 3, 'b']), I get one with quotes, but I don't want quotes.

Answer 1

In Python 2:

mylist = ['x', 3, 'b']
print '[%s]' % ', '.join(map(str, mylist))

In Python 3 (where print is a builtin function and not a syntax feature anymore):

mylist = ['x', 3, 'b']
print('[%s]' % ', '.join(map(str, mylist)))

Both return:

[x, 3, b]

This is using the map() function to call str for each element of mylist, creating a new list of strings that is then joined into one string with str.join(). Then, the % string formatting operator substitutes the string in instead of %s in "[%s]".

Answer 2

This is simple code, so if you are new you should understand it easily enough.

mylist = ["x", 3, "b"]
for items in mylist:
    print(items)

It prints all of them without quotes, like you wanted.

Answer 3

Using only print:

>>> l = ['x', 3, 'b']
>>> print(*l, sep='\n')
x
3
b
>>> print(*l, sep=', ')
x, 3, b

Answer 4

If you are using Python3:

print('[',end='');print(*L, sep=', ', end='');print(']')

Answer 5

Instead of using map, I'd recommend using a generator expression with the capability of join to accept an iterator:

def get_nice_string(list_or_iterator):
    return "[" + ", ".join( str(x) for x in list_or_iterator) + "]"

Here, join is a member function of the string class str. It takes one argument: a list (or iterator) of strings, then returns a new string with all of the elements concatenated by, in this case, ,.

Answer 6

You can delete all unwanted characters from a string using its translate() method with None for the table argument followed by a string containing the character(s) you want removed for its deletechars argument.

lst = ['x', 3, 'b']

print str(lst).translate(None, "'")

# [x, 3, b]

If you're using a version of Python before 2.6, you'll need to use the string module's translate() function instead because the ability to pass None as the table argument wasn't added until Python 2.6. Using it looks like this:

import string

print string.translate(str(lst), None, "'")

Using the string.translate() function will also work in 2.6+, so using it might be preferable.

Answer 7

Here's an interactive session showing some of the steps in @TokenMacGuy's one-liner. First he uses the map function to convert each item in the list to a string (actually, he's making a new list, not converting the items in the old list). Then he's using the string method join to combine those strings with ', ' between them. The rest is just string formatting, which is pretty straightforward. (Edit: this instance is straightforward; string formatting in general can be somewhat complex.)

Note that using join is a simple and efficient way to build up a string from several substrings, much more efficient than doing it by successively adding strings to strings, which involves a lot of copying behind the scenes.

>>> mylist = ['x', 3, 'b']
>>> m = map(str, mylist)
>>> m
['x', '3', 'b']
>>> j = ', '.join(m)
>>> j
'x, 3, b'

Answer 8

Using .format for string formatting,

mylist = ['x', 3, 'b']
print("[{0}]".format(', '.join(map(str, mylist))))

Output:

[x, 3, b]

---

Explanation:

1. map is used to map each element of the list to string type.
2. The elements are joined together into a string with , as separator.
3. We use [ and ] in the print statement to show the list braces.

Reference: .format for string formatting PEP-3101

Answer 9

I was inspired by @AniMenon to write a pythonic more general solution.

mylist = ['x', 3, 'b']
print('[{}]'.format(', '.join(map('{}'.format, mylist))))

It only uses the format method. No trace of str, and it allows for the fine tuning of the elements format. For example, if you have float numbers as elements of the list, you can adjust their format, by adding a conversion specifier, in this case :.2f

mylist = [1.8493849, -6.329323, 4000.21222111]
print("[{}]".format(', '.join(map('{:.2f}'.format, mylist))))

The output is quite decent:

[1.85, -6.33, 4000.21]