convert list into string in python3

Question

Right now I am having a list

>>> deints
[10, 10, 10, 50]

I want to print it as 10.10.10.50. I made it as

Method 1

>>> print(str(deints[0])+'.'+str(deints[1])+'.'+str(deints[2])+'.'+str(deints[3]))
10.10.10.50

Are there any other ways we can acheivie this ?

Thank you

• Sede: print('.'.join(map(str, deints)))
• PyNEwbie: Possible duplicate of Converting a list to a string

Answer 1

You can do it with:

print('.'.join(str(x) for x in deints))

• Raja G: Thank you. Even though it is for python 2 I understood what I need to add to make it run in python 3

Answer 2

This is very simple. Take a look at str.join

print '.'.join([str(a) for a in deints])

---

Citation from the docs:

str.join(iterable)

Return a string which is the concatenation of the strings in the iterable iterable. The separator between elements is the string providing this method.

• Raja G: Can you explain please
• ForceBru: @Raja, this just concatenates all the elements inserting a period as a separator. You also need to convert each number to a string first.

Answer 3

You can use the join method on strings and have to convert the data in the list to strings first.

>>> '.'.join(map(str, deints))
'10.10.10.50'

join takes the string as a delimiter and concatenates the content of the list with this delimiter between each element.

• ForceBru: I'm not the down voter, but the problem here is that you need to '.'.join
• Matthias: Ah, right. But the abstraction should have been possible. It's funny, first time I answered this kind of question I got positive reputation ...
• Raja G: I need an integer not again in string. I have found what I need. Thank you
• Matthias: @Raja: The list is unaltered. map applies the call to str to the list and gives back the changed result (an iterable with strings) which is then used by join.

Answer 4

Obviously str.join() is the shortest way

'.'.join(map(str, deints))

or if you dislike map() use a list comprehension

'.'.join([str(x) for x in deints])

you could also do it manually, using * to convert the deints list into a series of function arguments

'{}.{}.{}.{}'.format(*deints)

or use functools.reduce and a lambda function

reduce(lambda y, z: '{}.{}'.format(y, z), x)

All return

'10.10.10.50'

Answer 5

Just a non-join solution.

>>> print(*deints, sep='.')
10.10.10.50

Answer 6

Convert them into strings, then join them. Try using:

".".join([str(x) for x in x])