The strip command does not work on a list that contains other lists.
E.g. [ “ one”,”two “,[ “four”,”five”]]
I want to print out the words without the [].
Is there a way to get around this?
It turns out I was using append to build a list from other lists. I have now used the extend command instead to build the list which dispenses with the [] brackets, so printing is now straightforward. The extend command was not one I was familiar with before.
It's not entirely clear what you are trying to achieve, but if you just want to have a string representation without square brackets, you can do this:
l = [..., [...]]
s =str(l).replace("[", ""). replace("]", "")
print(s)
Mind that this can be misleading in many cases, because a list element that is a list of list looks like multiple string-type list elements. So you might want to reconsider why you want to do that.
Your [ and ] are forming a list, in your example a nested list.
You could use a recursive function to print it all out:
lst = [' one', 'two ', ['four', 'five']]
def recursive_print(my_list):
for item in my_list:
if isinstance(item, list):
recursive_print(item)
else:
print(item.strip())
recursive_print(lst)
This yields
one
two
four
five
If you want to collect the stripped words, you could use
def recursive_yield(my_list):
for item in my_list:
if isinstance(item, list):
yield from recursive_yield(item)
else:
yield item.strip()
flattened = [word for word in recursive_yield(lst)]
print(flattened)
Which would yield
['one', 'two', 'four', 'five']
Both functions will work for arbitrarily nested lists.
l=[ "one","two",[ "four","five"]]
for i in l:
if(isinstance(i,list)):
for j in i:
print(j,end=' ');#sep by spaces
else:
print(i,end=' ');
output:
one two four five
