Question

你好，我有一个问题，我正在努力解决。

定义一个名为clean_slate的方法，该方法将采用表示记录的String和另一个表示名称作为参数的String。该方法返回一个表示原始记录的字符串，但该名称的每个实例都被替换为两倍于名称包含字符的X＆＃39; s。如果传递了全名，它还将替换仅使用第一个或最后一个名称与全名包含字符的X的实例。您的方法不应区分大小写。假设输入始终是String。如果找不到姓名，则返回原始记录。

到目前为止，我有：

def clean_slate(record, name)
  name = name.downcase
  split_name = name.downcase.split(" ") #array
  change_name = split_name.map do |letter|
    letter = "X" * 2 * name.length
  end
  record_words = record.downcase.split(" ") #array
  new_record = record_words.map do |word|
   if word.include?(split_name[0] || split_name[1])
     word = change_name
   else 
     word 
   end
  end
 new_record.join(" ")
end

clean_slate("Megan Fox is a talented actress. megan knows sulfoxide is a chemical compound containing a sulfinyl (SO) functional group attached to two carbon atoms. It is a polar functional group. Ms. fox has many talents.", "Megan Fox")

我将返回record，但我没有在record编辑中获得姓氏。有谁知道如何解决这个问题？

"XXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXX fox is a talented actress. XXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXX knows sulfoxide is a chemical compound containing a sulfinyl (so) functional group attached to two carbon atoms. it is a polar functional group. ms. fox has many talents."

Answer 1

在record.split.map内，您需要检查传递的record.downcase是否包含split的第一个或最后一个值，但同时检查word是否new_record = record.split.map do |word| if record.downcase.include?(split_name[0] || split_name[1]) && word.downcase == split_name[0] || word.downcase == split_name[1] ...在块中等于这两个中的任何一个，所以它将类似于：

map

此外，如果您使用split_name而不是X来分配word = change_name的数量，那么它会返回一个数组，当您执行['XXXXX']时，它需要word = change_name.first数组中的值，如change_name.first（例如），因此您需要访问内部唯一的一个值：

def clean_slate(record, name)
  name = name.downcase
  split_name = name.downcase.split(" ")
  change_name = split_name.map do |letter|
    letter = 'X' * 2 * name.length
  end
  new_record = record.split.map do |word|
    if record.downcase.include?(split_name[0] || split_name[1]) &&
       word.downcase == split_name[0] || word.downcase == split_name[1]
      change_name.first
    else
      word
    end
  end
  new_record.join(" ")
end

由于不需要初始化未在块内使用的变量，因此只需split(" ")。

这样就可以了：

split

但它有一些你可以改进的东西：

letter = 'X' * 2 * name.length可以缩短为'XX' * name.length

change_name = 'XX' * name.length可以是change_name.first和

change_name无需使用地图，所以

join(" ")更改为if

record.map更喜欢单引号而不是双引号

您需要一种方法来简化[0]内的[-1]语句，但您仍然不知道该名称是否只有两个单词才能使用def clean_slate(record, name) record.gsub(/\b#{name.split.join('|')}\b/i, 'XX' * name.size) end和{{ 1}}。

您可以查看String#gsub。

#include<stdio.h>
#include<conio.h>

void main() 
{
    int a[]={1,2,3,4,5,6};
    int num,i;
    for(i=6;i>0;i--)
    {
        a[i]=a[i-1];
    }
    printf("enter any no, ");
    scanf("%d",&num); // input 12
    a[0]=num;
    for(i=0;i<=6;i++)
    {
        printf("\n\n%d",a[i]);
    }
    getch();
}

Answer 2

这是实现这一目标的更简洁方法。只需将record中的每个单词与name进行比较，然后替换X。

def clean_slate record, name
  names = name.downcase.split
  new_record = []
  record.split.each do |word|
    new_record << word
    if names.include?(word.downcase)
      new_record[-1] = "X" * 2 * name.length
    end
  end
  new_record.join(' ')
end

结果：

XXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXX is a talented actress. XXXXXXXXXXXXXXXXXX knows sulfoxide is a chemical compound containing a sulfinyl (SO) functional group attached to two carbon atoms. It is a polar functional group. Ms. XXXXXXXXXXXXXXXXXX has many talents.

Answer 3

如果我理解你的任务，我认为你可以做类似的解决方案：

def clean_slate(record, name)
  name = name.split.map(&:downcase)
  record = record.split

  record.map do |word|
    if name.include? word.downcase
      word = 'X' * 2 * name.join.length
    else
      word
    end
  end.join(' ')
end

#> clean_slate("Megan Fox is a talented actress. megan knows sulfoxide is a chemical compound containing a sulfinyl (SO) functional group attached to two carbon atoms. It is a polar functional group. Ms. fox has many talents.", "Megan Fox")
#=> "XXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXX is a talented actress. XXXXXXXXXXXXXXXX knows sulfoxide is a chemical compound containing a sulfinyl (SO) functional group attached to two carbon atoms. It is a polar functional group. Ms. XXXXXXXXXXXXXXXX has many talents."

用另一个的修改实例重写一个数组

3 个答案: