连续的字符序列 [英] Consecutive Character Sequences

问题描述

有没有办法[有效]迭代一系列字符来找到N [或更多]连续的等价字符？


所以，例如，字符串taaypiqee88adbbba ;如果在函数调用的参数

中提供的（连续字符数）为2或3，则返回1，因为a，e，8和b表示a，e，8和b。重复2到3次。


感谢您的帮助。

W. Brunswick。

Is there any way to [efficiently] iterate through a sequence of characters to find N [or more] consecutive equivalent characters?

So, for example, the string "taaypiqee88adbbba" would return 1 if the number (of consequtive characters) supplied in the parameters
of the function call was 2 or 3, because "a", "e", 8, and "b" is repeated 2 or 3 times.

Thanks for any assistance.
W. Brunswick.

推荐答案

Walter Brunswick写道：

Walter Brunswick wrote:

有没有办法[有效]迭代一系列字符来找到N [或更多]连续的等价字符？
中提供的（连续字符数）是2或3，则返回1，因为a，e，8和b表示a，e，8和b。重复2到3次。

Is there any way to [efficiently] iterate through a sequence of characters to find N [or more] consecutive equivalent characters?

So, for example, the string "taaypiqee88adbbba" would return 1 if the number (of consequtive characters) supplied in the parameters
of the function call was 2 or 3, because "a", "e", 8, and "b" is repeated 2 or 3 times.

为什么会返回1？这是代替True，还是代表一个

项目数，或者列表中某些东西的位置，或者是什么？


-Peter

Why would it return 1? Is that instead of True, or does it represent a
count of items, or the position of something in the list, or what?

-Peter

Walter Brunswick写道：

Walter Brunswick wrote:

有没有办法[有效]迭代序列找到N [或更多]连续等效字符的字符？

因此，例如，字符串taaypiqee88adbbba如果在函数调用的参数
中提供的（连续字符数）是2或3，则返回1，因为a，e，8和b表示a，e，8和b。重复2到3次。

感谢您的帮助。
W. Brunswick。

Is there any way to [efficiently] iterate through a sequence of characters to find N [or more] consecutive equivalent characters?

So, for example, the string "taaypiqee88adbbba" would return 1 if the number (of consequtive characters) supplied in the parameters
of the function call was 2 or 3, because "a", "e", 8, and "b" is repeated 2 or 3 times.

Thanks for any assistance.
W. Brunswick.

def rep（n）：

＃当前重复次数

the_rep = 1

＃检查前一个字符

last_c =''''

＃重复历史

max_rep = {}

for c in s：

＃找到重复的字符？

如果c == last_c：

＃计算连续多少重复

the_rep + = 1

#repethition（如果有的话）结束，保存以前的代表数

否则：

＃之前发生了这个计数吗？

如果max_rep.has_key（the_rep）：

＃如果是这样，跟踪它有多少次

max_rep [the_rep] + = 1

＃否则，将此重复计数添加到历史记录中

else：

max_rep [the_rep] = 1

#reset rep count to looking next block

the_rep = 1

＃保存当前字符以与下一个字符进行比较

last_c = c

＃检查字符串中的最后一个字符是不是块的一部分

如果max_rep.has_key（the_rep）：

max_rep [the_rep] + = 1

else：

max_rep [the_rep] = 1

＃finally，我们要求的块大小是否曾经发生过？

if max_rep.has_key（n）：

返回1

否则：

返回0


s = ''taaypiqee88adbbba''


for i in range（9）：

print rep（i），

"" ;"


0 1 1 1 0 0 0 0 0


"""

def rep(n):
# current repetition count
the_rep = 1
# previous character inspected
last_c = ''''
# history of repetions
max_rep = {}
for c in s:
# duplicate character found?
if c==last_c:
# count how many consecutive dups
the_rep += 1
# repetition (if any) ended, save previous rep count
else:
# has this count occured before?
if max_rep.has_key(the_rep):
# if so, track how many times it has
max_rep[the_rep] += 1
# otherwise, add this rep count to history
else:
max_rep[the_rep] = 1
# reset rep count to look for next block
the_rep = 1
# save current character to compare to next character
last_c = c
# check that last character in string wasn''t part of a block
if max_rep.has_key(the_rep):
max_rep[the_rep] += 1
else:
max_rep[the_rep] = 1
# finally, did the block size we asked for ever occur?
if max_rep.has_key(n):
return 1
else:
return 0

s = ''taaypiqee88adbbba''

for i in range(9):
print rep(i),
"""

0 1 1 1 0 0 0 0 0

"""

" Walter Brunswick" < WA ************* @ sympatico.ca>写道：

"Walter Brunswick" <wa*************@sympatico.ca> wrote:

有没有办法[有效]迭代一系列字符来找到N [或更多]
连续等效字符？
所以，对于例如，字符串taaypiqee88adbbba如果在函数调用的参数中提供的数字（连续的
个字符）是2或3，则返回1，因为a，e，8和b表示a，e，8和b。重复2到3次。

感谢您的帮助。
W. Brunswick。

Is there any way to [efficiently] iterate through a sequence of characters to find N [or more] consecutive equivalent characters?
So, for example, the string "taaypiqee88adbbba" would return 1 if the number (of consequtive characters) supplied in the parameters of the function call was 2 or 3, because "a", "e", 8, and "b" is repeated 2 or 3 times.

Thanks for any assistance.
W. Brunswick.

如果你在2.4，使用itertools.groupby：


导入itertools吧


def hasConsequent（aString，minConsequent）：

为_，group in it.groupby（aString）：

if len（list（group））> = minConsequent：

return true

返回False

George

If you''re in 2.4, use itertools.groupby:

import itertools as it

def hasConsequent(aString, minConsequent):
for _,group in it.groupby(aString):
if len(list(group)) >= minConsequent:
return True
return False
George

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