在Python中以特定模式打印字母

Question

我有以下字符串,我把它分开:

>>> st = '%2g%k%3p'
>>> l = filter(None, st.split('%'))
>>> print l
['2g', 'k', '3p']

现在我要打印两次g字母,k字母一次和p字母三次:

ggkppp

这怎么可能？

Answer 1

您可以使用generatorwith isdigit()来检查您的第一个符号是否为数字,然后返回具有适当计数的后续字符串.然后你可以join用来得到你的输出:

''.join(i[1:]*int(i[0]) if i[0].isdigit() else i for i in l)

示范:

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'

编辑

re当第一个数字有几个数字时使用模块:

''.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)

例:

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

您的输入如下:

st = '%2g_%3k%3p'

你可以_用空字符串替换,然后_如果列表中的工作以_符号结尾则添加到结尾:

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)

输出:

'gg_kkkppp'

EDIT3

没有re模块的解决方案,但通常的循环工作2位数.你可以定义函数:

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

然后将它们用作:

l = ['12g_', '3k', '3p']

print(collect(l))
gggggggggggg_kkkppp

Answer 2

单行正则表达方式:

>>> 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+)正则表达式匹配所有%并且在任何单词字符进入一个组之前存在零或多个数字,并且将后面的单词字符存储到另一个组中.在更换时,所有匹配的字符应替换为更换部件中给出的值.所以这应该是松散的%性格.

要么

>>> re.sub(r'%(\d*)(\w+)', lambda m: int(m.group(1))*m.group(2) if m.group(1) else m.group(2), st)
'ggkppp'

Answer 3

假设您始终打印单个字母,但前面的数字可能比基数10中的单个数字长.

seq = ['2g', 'k', '3p']
result = ''.join(int(s[:-1] or 1) * s[-1] for s in seq)
assert result == "ggkppp"

Answer 4

现在晚了但是准备好了

另一种方法是定义你的函数,它将nC转换为CCCC ... C(ntimes),然后将它传递给a map,将它应用于l来自splitover 的列表的每个元素%,最后join它们全部,如下所示:

>>> 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'

或者,如果您想将所有这些放在一个单一的函数定义中:

>>> 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

编辑:

如果存在_OP不希望重复此角色的地方,那么我认为最好的方法就是这样re.sub,它会让事情变得更容易,这样:

>>> 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'