#input
my_string = 'abcdefgABCDEFGHIJKLMNOP'
how would one extract all the UPPER from a string?
#output
my_upper = 'ABCDEFGHIJKLMNOP'
#input
my_string = 'abcdefgABCDEFGHIJKLMNOP'
how would one extract all the UPPER from a string?
#output
my_upper = 'ABCDEFGHIJKLMNOP'
Using list comprehension:
>>> s = 'abcdefgABCDEFGHIJKLMNOP'
>>> ''.join([c for c in s if c.isupper()])
'ABCDEFGHIJKLMNOP'
Using generator expression:
>>> ''.join(c for c in s if c.isupper())
'ABCDEFGHIJKLMNOP
You can also do it using regular expressions:
>>> re.sub('[^A-Z]', '', s)
'ABCDEFGHIJKLMNOP'
import string
s = 'abcdefgABCDEFGHIJKLMNOP'
s.translate(None,string.ascii_lowercase)
string.translate(s, table[, deletechars]) function will delete all characters from the string that are in deletechars, a list of characters. Then, the string will be translated using table (we are not using it in this case).
To remove only the lower case letters, you need to pass string.ascii_lowercase as the list of letters to be deleted.
The table
is None because when the table is None
, only the character deletion step will be performed.
Higher order functions to the rescue!
filter(str.isupper, "abcdefgABCDEFGHIJKLMNOP")
EDIT: In case you don't know what filter does: filter takes a function and an iterable, and then applies the function to every element in the iterable. It keeps all of the values that return true and throws out all of the rest. Therefore, this will return "ABCDEFGHIJKLMNOP".
or use regex ... this is an easy answer
import re
print ''.join(re.findall('[A-Z]+',my_string))
just for comparison
In [6]: %timeit filter(str.isupper,my_list)
1000 loops, best of 3: 774 us per loop
In [7]: %timeit ''.join(re.findall('[A-Z]+',my_list))
1000 loops, best of 3: 563 us per loop
In [8]: %timeit re.sub('[^A-Z]', '', my_list)
1000 loops, best of 3: 869 us per loop
In [10]: %timeit ''.join(c for c in my_list if c.isupper())
1000 loops, best of 3: 1.05 ms per loop
so this join plus findall is the fastest method (per ipython %timeit (python 2.6)) , using a 10000 character long identical string
edit: Or not
In [12]: %timeit my_list.translate(None,string.ascii_lowercase)
10000 loops, best of 3: 51.6 us per loop
You could use a more functional approach
>>> s = 'abcdefgABCDEFGHIJKLMNOP'
>>> filter(str.isupper, s)
'ABCDEFGHIJKLMNOP'
here you go:
my_string = 'abcdefgABCDEFGHIJKLMNOP'
cleanChar = ''
for char in my_string:
if char in 'ABCDEFGHIJKLMNOPQRSTUVWXYZ':
cleanChar = cleanChar + char
newChar = cleanChar
print(" {}".format(newChar))