Python string prints as [u'String']


This will surely be an easy one but it is really bugging me.

I have a script that reads in a webpage and uses Beautiful Soup to parse it. From the soup I extract all the links as my final goal is to print out the link.contents.

All of the text that I am parsing is ASCII. I know that Python treats strings as unicode, and I am sure this is very handy, just of no use in my wee script.

Every time I go to print out a variable that holds 'String' I get [u'String'] printed to the screen. Is there a simple way of getting this back into just ascii or should I write a regex to strip it?

4/14/2016 11:21:46 AM

Accepted Answer

[u'ABC'] would be a one-element list of unicode strings. Beautiful Soup always produces Unicode. So you need to convert the list to a single unicode string, and then convert that to ASCII.

I don't know exaxtly how you got the one-element lists; the contents member would be a list of strings and tags, which is apparently not what you have. Assuming that you really always get a list with a single element, and that your test is really only ASCII you would use this:


However, please double-check that your data is really ASCII. This is pretty rare. Much more likely it's latin-1 or utf-8.



Or you ask Beautiful Soup what the original encoding was and get it back in this encoding:

3/1/2009 11:40:01 AM

You probably have a list containing one unicode string. The repr of this is [u'String'].

You can convert this to a list of byte strings using any variation of the following:

# Functional style.
print map(lambda x: x.encode('ascii'), my_list)

# List comprehension.
print [x.encode('ascii') for x in my_list]

# Interesting if my_list may be a tuple or a string.
print type(my_list)(x.encode('ascii') for x in my_list)

# What do I care about the brackets anyway?
print ', '.join(repr(x.encode('ascii')) for x in my_list)

# That's actually not a good way of doing it.
print ' '.join(repr(x).lstrip('u')[1:-1] for x in my_list)

Licensed under: CC-BY-SA with attribution
Not affiliated with: Stack Overflow