I have the follwing string and I split it:
>>> st = '%2g%k%3p'
>>> l = filter(None, st.split('%'))
>>> print l
['2g', 'k', '3p']
Now I want to print the g letter two times, the k letter one time and the p letter three times:
ggkppp
How is it possible?
You could use generator with isdigit() to check wheter your first symbol is digit or not and then return following string with appropriate count. Then you could use join to get your output:
''.join(i[1:]*int(i[0]) if i[0].isdigit() else i for i in l)
Demonstration:
In [70]: [i[1:]*int(i[0]) if i[0].isdigit() else i for i in l ]
Out[70]: ['gg', 'k', 'ppp']
In [71]: ''.join(i[1:]*int(i[0]) if i[0].isdigit() else i for i in l)
Out[71]: 'ggkppp'
EDIT
Using re module when first number is with several digits:
''.join(re.search('(\d+)(\w+)', i).group(2)*int(re.search('(\d+)(\w+)', i).group(1)) if re.search('(\d+)(\w+)', i) else i for i in l)
Example:
In [144]: l = ['12g', '2kd', 'h', '3p']
In [145]: ''.join(re.search('(\d+)(\w+)', i).group(2)*int(re.search('(\d+)(\w+)', i).group(1)) if re.search('(\d+)(\w+)', i) else i for i in l)
Out[145]: 'ggggggggggggkdkdhppp'
EDIT2
For your input like:
st = '%2g_%3k%3p'
You could replace _ with empty string and then add _ to the end if the work from list endswith the _ symbol:
st = '%2g_%3k%3p'
l = list(filter(None, st.split('%')))
''.join((re.search('(\d+)(\w+)', i).group(2)*int(re.search('(\d+)(\w+)', i).group(1))).replace("_", "") + '_' * i.endswith('_') if re.search('(\d+)(\w+)', i) else i for i in l)
Output:
'gg_kkkppp'
EDIT3
Solution without re module but with usual loops working for 2 digits. You could define functions:
def add_str(ind, st):
if not st.endswith('_'):
return st[ind:] * int(st[:ind])
else:
return st[ind:-1] * int(st[:ind]) + '_'
def collect(l):
final_str = ''
for i in l:
if i[0].isdigit():
if i[1].isdigit():
final_str += add_str(2, i)
else:
final_str += add_str(1, i)
else:
final_str += i
return final_str
And then use them as:
l = ['12g_', '3k', '3p']
print(collect(l))
gggggggggggg_kkkppp
''.join(i[1:]*int(i[0]) if i[0].isdigit() else i for i in ['12g', 'k', '3p']) = '2gkppp'
st = '%2g_%3k%3p' so the result is:'g_g_kkkppp' but I want this result: 'gg_kkkppp'
_ sign?One-liner Regex way:
>>> import re
>>> st = '%2g%k%3p'
>>> re.sub(r'%|(\d*)(\w+)', lambda m: int(m.group(1))*m.group(2) if m.group(1) else m.group(2), st)
'ggkppp'
%|(\d*)(\w+) regex matches all % and captures zero or moredigit present before any word character into one group and the following word characters into another group. On replacement all the matched chars should be replaced with the value given in the replacement part. So this should loose % character.
or
>>> re.sub(r'%(\d*)(\w+)', lambda m: int(m.group(1))*m.group(2) if m.group(1) else m.group(2), st)
'ggkppp'
\b ?...because I think it's not necessary?re.sub(r'%(\d*)(\w*)', '-RE-',st) ---> '-RE--RE--RE-'
r'%(\d*)([a-z]+)'Assumes you are always printing single letter, but preceding number may be longer than single digit in base 10.
seq = ['2g', 'k', '3p']
result = ''.join(int(s[:-1] or 1) * s[-1] for s in seq)
assert result == "ggkppp"
result = ''.join([int(s[:-1] or 1) * s[-1] for s in st.split('%') if s]). You can give .join a generator expression but it's actually more efficient to give it a list comp. .join has to parse its input twice: the 1st pass determines the size of the output string, the 2nd pass builds the output. So if you give .join a gen exp it has to run the generator and build a list from it before it can start doing the actual joining.Another way, is to define your function which converts nC into CCCC...C (ntimes), then pass it to a map to apply it on every element of the list l coming from the split over %, the finally join them all, as follows:
>>> def f(s):
x = 0
if s:
if len(s) == 1:
out = s
else:
for i in s:
if i.isdigit():
x = x*10 + int(i)
out = x*s[-1]
else:
out = ''
return out
>>> st
'%4g%10k%p'
>>> ''.join(map(f, st.split('%')))
'ggggkkkkkkkkkkp'
>>> st = '%2g%k%3p'
>>> ''.join(map(f, st.split('%')))
'ggkppp'
Or if you want to put all of these into one single function definition:
>>> def f(s):
out = ''
if s:
l = filter(None, s.split('%'))
for item in l:
x = 0
if len(item) == 1:
repl = item
else:
for c in item:
if c.isdigit():
x = x*10 + int(c)
repl = x*item[-1]
out += repl
return out
>>> st
'%2g%k%3p'
>>> f(st)
'ggkppp'
>>>
>>> st = '%4g%10k%p'
>>>
>>> f(st)
'ggggkkkkkkkkkkp'
>>> st = '%4g%101k%2p'
>>> f(st)
'ggggkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkpp'
>>> len(f(st))
107
EDIT :
In case of the presence of _ where the OP does not want this character to be repeated, then the best way in my opinion is to go with re.sub, it will make things easier, this way:
>>> def f(s):
pat = re.compile(r'%(\d*)([a-zA-Z]+)')
out = pat.sub(lambda m:int(m.group(1))*m.group(2) if m.group(1) else m.group(2), s)
return out
>>> st = '%4g_%12k%p__%m'
>>> f(st)
'gggg_kkkkkkkkkkkkp__m'
gg_kkkkkkkkkkkkkppp ?But if I say st = %2g_%12k%3p result is not gg_kkkkkkkkkkkkkppp` or even when st = %2g___%12k%3p result must be gg___kkkkkkkkkkkkkpppLoop the list, check first entry for number, and then append the second digit onwards:
string=''
l = ['2g', 'k', '3p']
for entry in l:
if len(entry) ==1:
string += (entry)
else:
number = int(entry[0])
for i in range(number):
string += (entry[1:])
string isn't a great variable name since it's the name of a standard Python module. Also, this answer won't work properly if a number has more than 1 digit.
%mean, what values for the number are valid, and what values for the "string to print" are valid? For example, what would%234mean? Would it mean "print34twice," or is it invalid since no letter follows, or something else? What about%55a5? Printa5fifty five times, maybe, or print5a5five times, or print5five times and then printa5? There are a ton of cases you haven't specified here.