我正在尝试实现的是一个将字符串递增一个字符的函数,例如:
'AAA' + 1 = 'AAB'
'AAZ' + 1 = 'ABA'
'ZZZ' + 1 = 'AAAA'
我已经为前两种情况实现了功能,但我想不出第三种情况的任何解决方案。
这是我的代码:
def new_sku(s):
s = s[::-1]
already_added = False
new_sku = str()
for i in s:
if not already_added:
if (i < 'Z'):
already_added = True
new_sku += chr((ord(i)+1)%65%26 + 65)
else:
new_sku += i
return new_sku[::-1]
有什么建议吗?
答案 0 :(得分:9)
如果您正在处理bijective numeration,那么您可能已经(或应该)具有转换为/从双射表示的功能;只需转换为整数,增加它,然后转换回来就会轻松得多:
def from_bijective(s, digits=string.ascii_uppercase):
return sum(len(digits) ** i * (digits.index(c) + 1)
for i, c in enumerate(reversed(s)))
def to_bijective(n, digits=string.ascii_uppercase):
result = []
while n > 0:
n, mod = divmod(n - 1, len(digits))
result += digits[mod]
return ''.join(reversed(result))
def new_sku(s):
return to_bijective(from_bijective(s) + 1)
答案 1 :(得分:3)
怎么样?
def new_sku(s):
s = s[::-1]
already_added = False
new_sku = str()
for i in s:
if not already_added:
if (i < 'Z'):
already_added = True
new_sku += chr((ord(i)+1)%65%26 + 65)
else:
new_sku += i
if not already_added: # carry still left?
new_sku += 'A'
return new_sku[::-1]
示例运行: -
$ python sku.py Z
AA
$ python sku.py ZZZ
AAAA
$ python sku.py AAA
AAB
$ python sku.py AAZ
ABA
答案 2 :(得分:3)
你必须想到'AAA','ZZZ',......作为你操纵价值的代表。
首先,解析值:
val = sum(pow(26, i) * (ord(v) - ord('A') + 1) for i, v in enumerate(value[::-1]))
然后,为它增加价值:
val = val + 1
最终值由:
给出res = ""
while val > 0:
val, n = divmod(val - 1, 26)
res = chr(n+ord('A')) + res
缺少零表示要求传递给divmod的值在每个回合处递减,我没有找到一种方法来处理列表解析。
而不是ord()
和chr()
,可以使用string.ascii_uppercase.index()
和string.ascii_uppercase[]
答案 3 :(得分:2)
你可以在这里使用一些递归:
def new_sku(s):
s = s[::-1]
new_s = ''
return expand(s.upper(), new_s)[::-1]
import string
chars = string.ascii_uppercase
def expand(s, new_s, carry_forward=True):
if not s:
new_s += 'A' if carry_forward else ''
return new_s
new_s += chars[(ord(s[0]) - ord('A') + carry_forward) % 26]
# Slice the first character, and expand rest of the string
if s[0] == 'Z':
return expand(s[1:], new_s, carry_forward)
else:
return expand(s[1:], new_s, False)
print new_sku('AAB')
print new_sku('AAZ')
print new_sku('ZZZ')
print new_sku('aab')
print new_sku('aaz')
print new_sku('zzz')
输出:
AAC
ABA
AAAA
AAC
ABA
AAAA
答案 4 :(得分:1)
我会像携带底座26一样加入它。
所以从字符串的右边开始,添加1.如果它到达Z,则换行到A并将下一个最左边的字符向上碰撞一个。如果最左边的字符到达Z,则在字符串的左侧添加A.
s = ["Z","Z","Z"]
done = 0
index = len(s) - 1
while done == 0:
if s[index] < "Z":
s[index] = chr(ord(s[index]) + 1)
done = 1
else:
s[index] = "A"
if index == 0:
s = ["A"] + s
done = 1
else:
index = index - 1
print s
答案 5 :(得分:1)
检查字符串是否全部为Z
,如果是,请将其替换为长度为len(s) + 1
的字符串,仅包含A
s:
if s == "Z" * len(s):
return "A" * (len(s) + 1)
答案 6 :(得分:1)
alp='ABCDEFGHIJKLMNOPQRSTUVWXYZA'
def rec(s):
if len(s)==0:return 'A'
last_letter=s[-1]
if last_letter=='Z':return rec(s[:-1])+'A'
return s[:-1]+alp[(alp.find(last_letter)+1)]
结果
>>> rec('AAA')
'AAB'
>>> rec('AAZ')
'ABA'
>>> rec('ZZZ')
'AAAA'
>>> rec('AZA')
'AZB'
答案 7 :(得分:1)
这个怎么样?作为处理字符串越来越长的一种简单方法,你可以在前面加上'@'并在它没有增加的情况下将其删除:
>>> def new_sku(s):
def increment(s):
if s.endswith('Z'):
return increment(s[:-1])+'A'
else:
return s[:-1]+chr(ord(s[-1])+1)
t = increment('@'+s)
return t.lstrip('@')
>>> new_sku('AAA')
'AAB'
>>> new_sku('AAZ')
'ABA'
>>> new_sku('ZZZ')
'AAAA'
如果递归令你担心,那么你可以按照你已经做过的方式展平它,但仍然使用添加和剥离的'@'字符。
答案 8 :(得分:0)
您可以使用for-else
循环:
from string import ascii_uppercase as au
def solve(strs):
lis = []
for i, c in enumerate(strs[::-1], 1):
ind = au.index(c) + 2
lis.append(au[(ind%26)-1])
if ind <= 26:
break
else:
# This will execute only if the for-loop didn't break.
lis.append('A')
return strs[:-1*i] + "".join(lis[::-1])
print solve('AAA')
print solve('AAZ')
print solve('ZZZ')
print solve('AZZZ')
print solve('ZYZZ')
print solve('ZYYZZ')
<强>输出:强>
AAB
ABA
AAAA
BAAA
ZZAA
ZYZAA
答案 9 :(得分:0)
我们可以看到完全有3个条件,你可以迭代字符串并处理其中一个条件。 您可以使用string.ascii_uppercase而不是chr和ord
import string
def add(s):
s = list(s)[::-1]
for index, char in enumerate(s):
if char != "Z":
s[index] = string.ascii_uppercase[string.ascii_uppercase.index(char) + 1]
return s[::-1]
elif char == "Z" and (index != len(s) - 1):
s[index] = "A"
elif char == "Z" and (index == len(s) - 1):
s[index] = "A"
return ["A"] + s[::-1]