Question

我有一个几乎完全空白的data.frame，但每行都有一个值。如何使用矢量化或其他r-locald方法将每行的内容合并为一个向量？

示例数据：

raw_data  <- structure(
    list(
        col1 = c("", "", "", "", ""),
        col2 = c("", "", "", "", ""),
        col3 = c("", "", "", "", ""),
        col4 = c("", "", "", "Millburn - Union", ""),
        col5 = c("", "", "Cranston (aka Garden City Center)", "",""),
        col6 = c("", "", "", "", ""),
        col7 = c("", "", "", "", ""),
        col8 = c("", "", "", "", "Colorado Blvd"), 
        col9 = c("", "", "", "", ""),
        col10 = c("", "", "", "", ""),
        col11 = c("Palo Alto",  "Castro (aka Market St)", "", "", "")
    ),
    .Names = c("col1", "col2", "col3", "col4", "col5", "col6", "col7", "col8", "col9", "col10", "col11"),
    row.names = c(5L, 4L, 3L, 2L, 1L),
    class = "data.frame"
)

这是我尝试但它失败了，因为它返回一个二维矩阵而不是所需的向量：

raw_data$test <-  apply(raw_data, MAR=1, FUN=paste0)

Answer 1

您可以使用单个索引操作非常简单地执行此操作：

raw_data[raw_data!='']

演示：

R> raw_data[raw_data!=''];
[1] "Millburn - Union"                  "Cranston (aka Garden City Center)" "Colorado Blvd"                     "Palo Alto"                         "Castro (aka Market St)"

如果您关心矢量顺序是从上到下（而不是从左到右，然后从上到下，这就是上面的操作），您可以转置输入data.frame：

R> t(raw_data)[t(raw_data)!=''];
[1] "Palo Alto"                         "Castro (aka Market St)"            "Cranston (aka Garden City Center)" "Millburn - Union"                  "Colorado Blvd"

Answer 2

你对apply的直觉是正确的。您只需将collapse参数传递给paste：

 apply( raw_data, 1, paste0, collapse = "" )
                                  5                                   4                                   3 
                        "Palo Alto"            "Castro (aka Market St)" "Cranston (aka Garden City Center)" 
                                  2                                   1 
                 "Millburn - Union"                     "Colorado Blvd"

Answer 3

在此示例中，每行只有一个元素不是''。这是另一种将paste与do.call

一起使用的方法

do.call(paste, c(raw_data, sep=''))
#[1] "Palo Alto"                         "Castro (aka Market St)"           
#[3] "Cranston (aka Garden City Center)" "Millburn - Union"                 
#[5] "Colorado Blvd"

假设，如果'raw_data'中每行有多个元素不是''，那么在这种情况下，最好使用sep=';' or，`。

raw_data[1,1] <- 'Millburn'
raw_data[1,3] <- 'Something'
gsub('^;+|;+$|(;);+', '\\1', do.call(paste, c(raw_data, sep=';')))
#[1] "Millburn;Something;Palo Alto"      "Castro (aka Market St)"     
#[3] "Cranston (aka Garden City Center)" "Millburn - Union"           
#[5] "Colorado Blvd"

apply获得与上面相同的结果

unname(apply(raw_data, 1, FUN=function(x) paste(x[x!=''],collapse=';')))
#[1] "Millburn;Something;Palo Alto"      "Castro (aka Market St)"    
#[3] "Cranston (aka Garden City Center)" "Millburn - Union"         
#[5] "Colorado Blvd"

使用apply或其他矢量化方法合并列的内容

3 个答案: