Python 遍历长度未知的数字的字母
好的。。。我需要遍历任意长度的字符串。因为我不知道怎么解释,我的意思是:Python 遍历长度未知的数字的字母,python,python-2.7,Python,Python 2.7,好的。。。我需要遍历任意长度的字符串。因为我不知道怎么解释,我的意思是: def zip(string1,string1): ... 当使用“a”和“ad”调用时,它将返回一个列表: >>>zip("a","ad") ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","aa","ab","ac","ad"]
def zip(string1,string1):
...
当使用“a”
和“ad”
调用时,它将返回一个列表:
>>>zip("a","ad")
["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","aa","ab","ac","ad"]
我尝试过使用
map(chr,range(ord('a'),ord('nb')+1))
,但我得到TypeError:ord()应该是一个字符,但找到了长度为2的字符串,我不知道从这里走到哪里。有什么想法吗?这是一个base-26数字系统,下面是我如何解决它的。而且zip
是一个python内置函数,最好不要重新定义它
def alphaToNumber(s):
r = 0
for x in s:
r *= 26
r += ord(x) - 96
return r
def numberToAlpha(n, result):
head = (n - 1) // 26
tail = chr((n - 1) % 26 + 97)
if head == 0:
return tail + result
else:
return numberToAlpha(head, tail + result)
def gen(start, end):
start_n = alphaToNumber(start)
end_n = alphaToNumber(end)
return [numberToAlpha(x, "") for x in range(start_n, end_n + 1)]
print(gen("a", "ad"))
# ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z', 'aa', 'ab', 'ac', 'ad']
这是一个base-26数字系统,下面是我将如何解决它。而且zip
是一个python内置函数,最好不要重新定义它
def alphaToNumber(s):
r = 0
for x in s:
r *= 26
r += ord(x) - 96
return r
def numberToAlpha(n, result):
head = (n - 1) // 26
tail = chr((n - 1) % 26 + 97)
if head == 0:
return tail + result
else:
return numberToAlpha(head, tail + result)
def gen(start, end):
start_n = alphaToNumber(start)
end_n = alphaToNumber(end)
return [numberToAlpha(x, "") for x in range(start_n, end_n + 1)]
print(gen("a", "ad"))
# ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z', 'aa', 'ab', 'ac', 'ad']
就是这样:
def zip_(start, end):
def __inc(s):
if not s:
return "a"
elif s[-1] != "z":
return s[:-1] + chr(ord(s[-1]) + 1)
else:
return __inc(s[:-1]) + "a"
s = start
yield s
while s != end:
s = __inc(s)
yield s
print list(zip_("a", "ad"))
几点意见:
不要使用zip
作为变量或函数的名称,因为它已经被保留
在解决方案中,zip
是一个生成器。我这样做是为了不在内存中保存太多数据。如果您需要一个精确的列表,只需像我在print语句中所做的那样转换它
如果参数错误,函数可能会进入无限循环。例如,如果您调用zip(“b”,“a”)
。但实际上,如果有必要的话,可以通过添加几行来轻松修复
就是这样:
def zip_(start, end):
def __inc(s):
if not s:
return "a"
elif s[-1] != "z":
return s[:-1] + chr(ord(s[-1]) + 1)
else:
return __inc(s[:-1]) + "a"
s = start
yield s
while s != end:
s = __inc(s)
yield s
print list(zip_("a", "ad"))
几点意见:
不要使用zip
作为变量或函数的名称,因为它已经被保留
在解决方案中,zip
是一个生成器。我这样做是为了不在内存中保存太多数据。如果您需要一个精确的列表,只需像我在print语句中所做的那样转换它
如果参数错误,函数可能会进入无限循环。例如,如果您调用zip(“b”,“a”)
。但实际上,如果有必要的话,可以通过添加几行来轻松修复
一开始我有点困惑,因为zip是一个实际的python方法。一开始我有点困惑,因为zip是一个实际的python方法。谢谢!这是完美的工作,我喜欢它是一个发电机太!谢谢这是完美的工作,我喜欢它是一个发电机太!