Python: most idiomatic way to convert None to empty string?

Question

What is the most idiomatic way to do the following?

def xstr(s):
    if s is None:
        return ''
    else:
        return s

s = xstr(a) + xstr(b)

update: I'm incorporating Tryptich's suggestion to use str(s), which makes this routine work for other types besides strings. I'm awfully impressed by Vinay Sajip's lambda suggestion, but I want to keep my code relatively simple.

def xstr(s):
    if s is None:
        return ''
    else:
        return str(s)

Answer 1

Probably the shortest would be str(s or '')

Because None is False, and "x or y" returns y if x is false. See Boolean Operators for a detailed explanation. It's short, but not very explicit.

Answer 2

def xstr(s):
    return '' if s is None else str(s)

Answer 3

If you actually want your function to behave like the str() built-in, but return an empty string when the argument is None, do this:

def xstr(s):
    if s is None:
        return ''
    return str(s)

Answer 4

If you know that the value will always either be a string or None:

xstr = lambda s: s or ""

print xstr("a") + xstr("b") # -> 'ab'
print xstr("a") + xstr(None) # -> 'a'
print xstr(None) + xstr("b") # -> 'b'
print xstr(None) + xstr(None) # -> ''

Answer 5

return s or '' will work just fine for your stated problem!

Answer 6

def xstr(s):
   return s or ""

Answer 7

UPDATE:

I mainly use this method now:

some_string = None
some_string or ''

If some_string was not NoneType, the or would short circuit there and return it, otherwise it returns the empty string.

OLD:

Max function worked in python 2.x but not in 3.x:

max(None, '')  # Returns blank
max("Hello", '') # Returns Hello

Answer 8

Functional way (one-liner)

xstr = lambda s: '' if s is None else s

Answer 9

A neat one-liner to do this building on some of the other answers:

s = (lambda v: v or '')(a) + (lambda v: v or '')(b)

or even just:

s = (a or '') + (b or '')

Answer 10

def xstr(s):
    return {None:''}.get(s, s)

Answer 11

Variation on the above if you need to be compatible with Python 2.4

xstr = lambda s: s is not None and s or ''

Answer 12

If it is about formatting strings, you can do the following:

from string import Formatter

class NoneAsEmptyFormatter(Formatter):
    def get_value(self, key, args, kwargs):
        v = super().get_value(key, args, kwargs)
        return '' if v is None else v

fmt = NoneAsEmptyFormatter()
s = fmt.format('{}{}', a, b)

Answer 13

def xstr(s):
    return s if s else ''

s = "%s%s" % (xstr(a), xstr(b))

Answer 14

We can always avoid type casting in scenarios explained below.

customer = "John"
name = str(customer)
if name is None
   print "Name is blank"
else: 
   print "Customer name : " + name

In the example above in case variable customer's value is None the it further gets casting while getting assigned to 'name'. The comparison in 'if' clause will always fail.

customer = "John" # even though its None still it will work properly.
name = customer
if name is None
   print "Name is blank"
else: 
   print "Customer name : " + str(name)

Above example will work properly. Such scenarios are very common when values are being fetched from URL, JSON or XML or even values need further type casting for any manipulation.

Answer 15

Use short circuit evaluation:

s = a or '' + b or ''

Since + is not a very good operation on strings, better use format strings:

s = "%s%s" % (a or '', b or '')

Answer 16

Same as Vinay Sajip's answer with type annotations, which precludes the needs to str() the result.

def xstr(s: Optional[str]) -> str:
    return '' if s is None else s

Answer 17

Use F string if you are using python v3.7

xstr = F"{s}"