Question

我在Python中尝试计算字符串中的子字符串。

string = "aaaaahehabdha" 

x = string.index("he") #5
y = string.rindex("ha") #11
z = y - x + 2 #lenght of substring 11 - 5 + 2 = 8

a = []
a.append(string.count("he", x, z)) #1
a.append(string.count("ha", x, z)) #count only once"ha" but in substring I have 2 x "ha"
print(a) # [1,1]

为什么string.count（“ha”，x，z）只计算一次“ha”？提前谢谢。

Answer 1

.count的第三个参数不是长度，而是非包含性结束索引。计算正确的终点：

z = y + 2

示例，包括评论示例，计算从he到ha的正确子字符串：

string = "aaaaahehabdha" 
x = string.index("he")
y = string.rindex("ha")
z = y + 2
print('substring:',string[x:z])

string = "aaaaahehabdhaaaaaaaaaaa" 
x = string.index("he")
y = string.rindex("ha")
z = y + 2
print('substring:',string[x:z])

输出：

substring: hehabdha
substring: hehabdha

Answer 2

如果您只想计算子字符串中“ha”的数量，您只需使用string.count("ha")，它将返回2.

Answer 3

您的参数public function buildForm(FormBuilderInterface $builder, array $options) { $builder ->add('firstname', TextType::class, ['attr' => ['class' => 'form-control']]) ->add('secondname', TextType::class, ['attr' => ['class' => 'form-control']]) ->add('email', EmailType::class, ['attr' => ['class' => 'form-control']]) ->add('password', PasswordType::class, ['attr' => ['class' => 'form-control']]) ->add('password_confirmation', PasswordType::class, [ 'label' => 'Confirm Password', 'attr' => ['class' => 'form-control'], 'mapped' =>false ]) ->add('Register', SubmitType::class, ['attr' => ['class' => 'btn btn-primary']]); }和x是在字符串中搜索位置的起始和结束索引，但起始索引始终是包含的，而结束索引始终是非包含的。您在z之后存储了要搜索的起始索引之后的长度，当您确实应该调用时

Python字符串分析，计算子字符串错误

3 个答案: