In Python, I can do:
list = ['a', 'b', 'c']
', '.join(list) # 'a, b, c'
However, if I have a list of objects and try to do the same thing:
class Obj:
def __str__(self):
return 'name'
list = [Obj(), Obj(), Obj()]
', '.join(list)
I get the following error:
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: sequence item 0: expected string, instance found
Is there any easy way? Or do I have to resort to a for-loop?
You could use a list comprehension or a generator expression instead:
', '.join([str(x) for x in list]) # list comprehension
', '.join(str(x) for x in list) # generator expression
map expression (below).
map solution.The built-in string constructor will automatically call obj.__str__:
''.join(map(str,list))
reduce was the one that was removed, because it usually left people guessing and thus wasn't "pythonic". map on the other hand is not an issue.
I know this is a super old post, but I think what is missed is overriding __repr__, so that __repr__ = __str__, which is the accepted answer of this question marked duplicate.
another solution is to override the join operator of the str class.
Let us define a new class my_string as follows
class my_string(str):
def join(self, l):
l_tmp = [str(x) for x in l]
return super(my_string, self).join(l_tmp)
Then you can do
class Obj:
def __str__(self):
return 'name'
list = [Obj(), Obj(), Obj()]
comma = my_string(',')
print comma.join(list)
and you get
name,name,name
BTW, by using list as variable name you are redefining the list class (keyword) ! Preferably use another identifier name.
Hope you'll find my answer useful.
Not str.join but you can also use str.format to join the objects into a single string, especially if __str__ method is defined and lst is not very long.
class Obj:
def __str__(self):
return "name"
lst = [Obj(), Obj(), Obj()]
"{}, {}, {}".format(*lst) # 'name, name, name'
If lst is very long and initializing specific number of placeholders is not feasible, then we can initialize a string of placeholders and format it afterwards (admittedly, very ugly though).
(len(lst)*"{}, ").format(*lst).rstrip(', ')