在数组中查找重复的字符串并修改它们答案

【问题标题】：Find Duplicate strings in an array and modify them在数组中查找重复的字符串并修改它们
【发布时间】：2023-03-20 20:10:01
【问题描述】：

我有一大堆字符串，这里是其中的一小部分：

let x = [
  'FireDisaster_03_', 
  'FireDisaster_03_', 
  'FireDisaster_03_', 
  'FireDisaster_05_', 
  'FireDisaster_05_', 
  'FireDisaster_07_', 
  'FireDisaster_07_', 
  'FireDisaster_07_', 
  'FireDisaster_07_', 
  'FireDisaster_08_', 
  'FireDisaster_08_', 
  'FireDisaster_08_'
]

我需要为每个重复的字符串添加序号。这样结果将如下所示：

[
  'FireDisaster_03_0', 
  'FireDisaster_03_1',  
  'FireDisaster_03_2', 
  'FireDisaster_05_0', 
  'FireDisaster_05_1', 
  'FireDisaster_07_0', 
  'FireDisaster_07_1', 
  'FireDisaster_07_2', 
  'FireDisaster_07_3', 
  'FireDisaster_08_0', 
  'FireDisaster_08_1', 
  'FireDisaster_08_2'
]

请帮我解决这个问题

谢谢

【问题讨论】：

用什么来做这个？
嗯，它更多的是机器学习目的，有很多图像文件但它们没有正确命名，我能够将它提升到我现在拥有的但无法为重复名称添加数字

标签： javascript arrays duplicates

【解决方案1】：

您可以为所见值获取一个对象并存储计数器。

这种方法在 seen 上使用闭包，带有一个 IIFE (immediately-invoked function expression) 和一个检查键是否存在于对象中的表达式。

在第一种情况下，它从属性中获取值并加一个；在第二种情况下，它将零作为分配给属性的值，同时该值用于与字符串连接。

let x = ['FireDisaster_03_', 'FireDisaster_03_', 'FireDisaster_03_', 'FireDisaster_05_', 'FireDisaster_05_', 'FireDisaster_07_', 'FireDisaster_07_', 'FireDisaster_07_', 'FireDisaster_07_', 'FireDisaster_08_', 'FireDisaster_08_', 'FireDisaster_08_'],
    result = x.map(
        (seen => v => v + (seen[v] = v in seen ? seen[v] + 1 : 0))
        ({})
    );

console.log(result);

【讨论】：

【解决方案2】：

您可以使用reduce。还在操作之前对数组进行排序以仅与前一个元素进行比较

let x = ['FireDisaster_03_', 'FireDisaster_03_', 'FireDisaster_03_', 'FireDisaster_05_', 'FireDisaster_05_', 'FireDisaster_07_', 'FireDisaster_07_', 'FireDisaster_07_', 'FireDisaster_07_', 'FireDisaster_08_', 'FireDisaster_08_', 'FireDisaster_08_'].sort();
let newStrArray = x.reduce((accumulator, currentVal, index) => {
  // for the first element of the array no changes need to be done
  if (index === 0) {
    accumulator.push(`${currentVal}${index}`);
  } else {
    // check if the current element is same as the previous element from
    // the main array
    const prevVal = x[index - 1];
    if (prevVal === currentVal) {
      // if same then get the previous element from accumulator array
      const getPrevVal = accumulator[index - 1];
      // get the last character , convert to number and increase it by 1
      const num = parseInt(getPrevVal.charAt(getPrevVal.length - 1), 10) + 1;
      // push value to accumulator array
      accumulator.push(`${currentVal}${num}`)
    } else {
      // if previous and current element are not same then 
      // push current element in accumulator array
      accumulator.push(`${currentVal}0`)
    }
  }


  return accumulator;
}, []);
console.log(newStrArray)

【讨论】：

【解决方案3】：

您可以像这样使用递归函数和.map()：

let x = ['FireDisaster_03_', 'FireDisaster_03_', 'FireDisaster_03_', 'FireDisaster_05_', 'FireDisaster_05_', 'FireDisaster_07_', 'FireDisaster_07_', 'FireDisaster_07_', 'FireDisaster_07_', 'FireDisaster_08_', 'FireDisaster_08_', 'FireDisaster_08_']

x = x.map((c, i, a) =>
  a.indexOf(c) === i ? c + '0' : f(a.indexOf(c), c, i, a.slice(1), 1)
);

function f(ac, c, i, a, s) {
  return a.indexOf(c) === i - s ? c + (s - ac) : f(ac, c, i, a.slice(1), s + 1);
}

console.log(x);

【讨论】：

【解决方案4】：

found = {}
x.forEach( (val, i) => {
    found[val] = (found[val]||0) +1;
    x[i] += found[val]-1;
})

如果你不介意从 1 开始，可以去掉最后一个 -1

【讨论】：