Question

我正在R。

我有两个tibbles / data.frames值，均由一列year索引。

第一个小标题df-包含n years的值（不一定唯一，但可以是唯一的），第二个小标题-df1 -包含每个year的第一个的乘法值（在这种情况下，仅unique(year)个值）。

我要做的是将第一矩阵的每个元素乘以第二矩阵的正确引用。

Year <- c("2010", "2010", "2011", "2011")
value <- c(1, 2, 3, 4)          
df <- as_tibble(value) %>% add_column(Year= Year)

Year1<- c("2010", "2011")
value1 <- c(100, 200)

df1 <- as_tibble(value1) %>% add_column(Year1= Year1)

此示例的结果应为以下内容（但可扩展为动态dim(df) / dim(df1)：

result <- c(100, 200, 600, 800)
res <- as_tibble(result) %>% add_column(Year = Year)

任何想法我该如何实现？非常感谢。

Answer 1

我们可以在“年份”列中加入processData，然后将“值”列相乘

function processData(data, json) {
  const combinators = ["OR", "AND"]

  function leafValue(item) {
    switch (item.operator) {
      case '=':
        return data[item.value] === item.expression
    }
    throw new Error(`unknown comparison operator ${item.operator}`)
  }

  function evaluate(json) {
    const evaluated = json.map(item => Array.isArray(item) ? evaluate(item) : item)
    while (evaluated.length >= 3) {
      const chunk = [evaluated[0], evaluated[1], evaluated[2]]
      const combinator = chunk.find(item => combinators.includes(item))
      const [A, B] = chunk.filter(item => !combinators.includes(item))
      const valueA = typeof A === 'boolean' ? A : leafValue(A)
      const valueB = typeof B === 'boolean' ? B : leafValue(B)
      const result = combinator === 'OR' ?
        (valueA || valueB) :
        (valueA && valueB)
      evaluated.splice(0, 3, result)
    }
    return evaluated[0]
  }

  return evaluate(json)
}

const json = {
  "truthy": [
    [
      "AND",
      {
        "operator": "=",
        "compareType": "text",
        "value": "fname",
        "expression": "John"
      },
      {
        "operator": "=",
        "compareType": "text",
        "value": "lname",
        "expression": "Doe"
      }
    ],
    "OR",
    {
      "operator": "=",
      "compareType": "boolean",
      "value": "Agree",
      "expression": true
    },
    "OR", [
      "AND",
      {
        "operator": "=",
        "compareType": "text",
        "value": "fname",
        "expression": "Jane"
      },
      {
        "operator": "=",
        "compareType": "text",
        "value": "lname",
        "expression": "Doe"
      }
    ]
  ]
}

const data = {
  fname: 'John',
  lname: 'Doe',
}

const result = processData(data, json.truthy)
console.log(result)

或者使用by连接并通过将“值”乘以第二个数据集（library(dplyr) df %>% left_join(df1, by = c('Year' = 'Year1')) %>% transmute(value = value.x * value.y, Year)）的“值”列来分配（data.table）“值”

:=

Answer 2

这是使用merge

的基本R选项

transform(
  merge(df, df1, by.x = "Year", by.y = "Year1"),
  value = value.x * value.y
)[names(df)]

给出

  value Year
1   100 2010
2   200 2010
3   600 2011
4   800 2011

将矩阵乘以包含在不同维度的第二个矩阵中的参考值

2 个答案: