关于一段程序的问题
题目的要求是每个找出str1中 的小写字母 且这个小写字母左右两边均有且仅有3个大写字母我的程序是这样的
def count(x):
list1 = []
str2 = ''
lenth = len(x)
for i in range(lenth - 3):
if x.islower() == True: #这里找出str1中的小写字母
if (x[(i - 3):i].isupper() == True) and (x[(i + 1):(i + 4)].isupper == True): #这里试图筛选出两边有3个大写字母的小写字母(似乎?问题出在了这里?)
if x.isupper() == False and x.isupper() == False: #这里剔除掉不仅仅只有3个大写字母的情况
list1.append(x)
continue
return list1
print(count(str1))
我百思不得其解 为什么我最后的list1是一个空列表
下面是str1
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qKWGc""" 本帖最后由 isdkz 于 2022-3-13 08:12 编辑
def count(x):
list1 = []
str2 = '' # 这个 str2 至始至终就没有用到,可以不要
lenth = len(x)
for i in range(lenth - 3): # 这里不能从 0 开始,因为 0 - 3 就为负数了,也不能到 lenth - 3 结束,range时左闭右开的,所以应该是到 length-2 结束
if x.islower() == True: # 这个 if 就没有被满足过,因为你用 x(x 是传进来的整段字符串) 来调用 islower 方法,只要 x 不全为小写,islower() 就不会为True,这是导致你的列表为 空 的根本原因
if (x[(i - 3):i].isupper() == True) and (x[(i + 1):(i + 4)].isupper == True): # 这里的最后一个 isupper 少了括号
if x.isupper() == False and x.isupper() == False:
list1.append(x) # 这里应该时 append(x),x 是传进来的整段字符串
continue # 这里本就在循环体的最后面,这个 continue 有点多余
return list1
print(count(str1))
故对你的代码修改如下:
def count(x):
list1 = []
# str2 = '' # 改了这里
lenth = len(x)
for i in range(3, lenth - 2): # 改了这里
if x.islower() == True: # 改了这里
if i - 3 == 0 and x.isupper() == False and (x[(i + 1):(i + 4)].isupper() == True): #注意这里,因为 i 字符有可能左边就三个字符或者右边就三个字符,比如字符串的第四个字符 i,左边就三个字符,如果 减4 的话就是倒数第一个字符了
list1.append(x)
elif i + 4 == lenth and x.isupper() == False and (x[(i - 3):i].isupper() == True): # 注意这里
list1.append(x)
elif (x[(i - 3):i].isupper() == True) and (x[(i + 1):(i + 4)].isupper() == True): # 改了这里
if x.isupper() == False and x.isupper() == False:
list1.append(x) # 改了这里
return list1
print(count(str1))
isdkz 发表于 2022-3-13 07:38
def count(x):
list1 = []
str2 = '' # 这个 str2 至始至终就没有用 ...
emmmmmmmmm大致看懂了 但是改了程序之后 打印出来的字符串还是与答案有差距 多了几个字母 答案应该是i love fishc 请问为啥会多出几个字母呢
本帖最后由 isdkz 于 2022-3-13 13:20 编辑
Jerry~Mouse 发表于 2022-3-13 13:04
emmmmmmmmm大致看懂了 但是改了程序之后 打印出来的字符串还是与答案有差距 多了几个字母 答案应该是i lo ...
因为小甲鱼的字符串跟你的不一样,你的粘贴上来的是带 换行 的,
>>> 'AB\n'.isupper() # 可以看到带着换行符也是会被判断为 True 的,你的字符串虽然没有 \n,但是三引号就是会把你换行的地方自动转为 换行符 \n
True
>>>
小甲鱼的没有换行,是连着的。
你的字符串用 replace 方法把换行去掉就跟小甲鱼的一样了,
replace用法:
str1.replace('\n', '') # 替换掉字符串中的换行符为空字符,即去掉换行符 isdkz 发表于 2022-3-13 13:17
因为小甲鱼的字符串跟你的不一样,你的粘贴上来的是带 换行 的,
这是不是也就意味着 有两个大写字母的换行处也会被判定为True,如果用replace的话,可以替换掉潜在的换行符,可以这么理解吗,只有三引号有潜在换行符吗 本帖最后由 isdkz 于 2022-3-13 13:57 编辑
Jerry~Mouse 发表于 2022-3-13 13:40
这是不是也就意味着 有两个大写字母的换行处也会被判定为True,如果用replace的话,可以替换掉潜在的换行 ...
嗯嗯,对的 isdkz 发表于 2022-3-13 13:51
嗯嗯,对的
明白了 谢谢谢谢 lenth = len(str1)
a=list(str1)
print(a)
for i in range(lenth-3):
if a.islower()==True :
if (a.isupper() == True)and(a.isupper() == True)and(a.isupper() == True)and(a.isupper() == True)and(a.isupper() == True)and(a.isupper() == True):
print(a+a+a+a+a+a+a)
测试成功
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