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我有三个文本文件。我想做一些如下所示的计算并写出结果。所有文本文件都包含 14 列 X1 到 X14 和 601 行。此代码基本上计算所有三个文件之间的相关性（例如，来自 ref 文件的 x1 和来自文件 sour1 的同一列 x1 等）并进行一些计算，然后返回结果。

ref= read.table("D:\\S_asc.txt", sep="",header=TRUE)
sour1 = read.table("D:\\sre.txt", sep="",header=TRUE) 
sour2= read.table("D:\\os_asc.txt", sep="",header=TRUE)
columns <- paste0("X", 1:13)
lapply(
    columns,
    function(column)
    {
        result1 <- cor(ref[[column]],sour2[[column]],use = "na.or.complete")
        # calculate using ref and sour1

        result2 <-  cor(ref[[column]],sour1[[column]],use = "na.or.complete")   
        # calculate using ref and sour2

    }
)

编写结果时出现错误：

write.table(result1,file = "foo.text")
Error in is.data.frame(x) : object 'result1' not found

我的文件样本：

"0" "1" "2" "3" "4" "5" "6" "7" "8" "9" "10" "11" "12" "13" "14"
"2" 0.189444444142282 0.317999997238318 0.166363635072202 0.301255410351666 0.471986238597248       0.307140744955515 0.411269603728529 0.262293613227113 0.265422601640182 0.196970316722191      0.123789306669182 0.291404434246697 0.322141939017999 0.43499182336885 0.327559809775301
"3" 0.149411764616768 0.191176471464774 0.180181817100807 0.422626723831494 0.496283823857852 0.327019521055943 0.441349119295007 0.286810015617714 0.24089.

        dput( head( ref ) ) 
          structure(list(X0 = c(0.0493461075417879, 0.0505162129561387, 
       0.0490028816673631, 0.049519460579677, 0.0496427079798826, 0.0506145435593228
       ), X1 = c(0.0262911450465596, 0.0249440745179165, 0.0240951294080254, 
      0.0256401352066387, 0.0258794329817874, 0.0247771567181071), 
        X2 = c(0.0426611743228151, 0.0374958657769128, 0.0306825108745032, 
         0.0393093163299325, 0.0398889994476742, 0.0435458658141826
         ), X3 = c(0.0954854469641096, 0.0833464493313783, 0.0790441582634745, 
         0.0818814458477848, 0.0772971938289144, 0.0818534809699011
        ), X4 = c(0.0933782886825547, 0.0968294484334356, 0.0981543002252867, 
         0.0969233013221434, 0.0933197039086707, 0.0913029715102785
         ), X5 = c(0.218212200747129, 0.220014530510402, 0.21109789420558, 
         0.204679954607016, 0.210967709615173, 0.20799168387628), 
       X6 = c(0.285405481705908, 0.284798691283541, 0.280772749846878, 
         0.285532926668878, 0.293501364933202, 0.296710869439616), 
        X7 = c(0.226218243796976, 0.216760771202585, 0.205519652752883, 
        0.206054283066245, 0.211842009508557, 0.214360803181363), 
       X8 = c(0.146648210899044, 0.144789395382814, 0.142059144367336, 
        0.141852209487122, 0.140823495101071, 0.140405076285803), 
        X9 = c(0.115716572518044, 0.113825119638324, 0.111749156097298, 
        0.111175507359272, 0.11094649408726, 0.111155298251533), 
         X10 = c(0.0675501818197432, 0.0636083677843573, 0.0605178392071445, 
           0.0549028963149205, 0.0600036757169919, 0.0625667586974292
         ), X11 = c(0.069120070466305, 0.0662362222295113, 0.0710726619403152, 
         0.0653345812188053, 0.0612221754441228, 0.0584222641456072
            ), X12 = c(0.281314574594234, 0.279000957729072, 0.276184085574267, 
            0.274121716868717, 0.271324697416127, 0.270268668362535), 
              X13 = c(0.364434947521643, 0.369388875815867, 0.372250445014838, 
          0.372470394613093, 0.372814994985087, 0.369151832740263), 
                X14 = c(0.0124844383491671, 0.0125512696973245, 0.0125538467586845, 
                   0.012599469429366, 0.0125682867660489, 0.0126515616426303
                   )), .Names = c("X0", "X1", "X2", "X3", "X4", "X5", "X6", 
                     "X7", "X8", "X9", "X10", "X11", "X12", "X13", "X14"), row.names = c("2", 
                     "3", "4", "5", "6", "7"), class = "data.frame")
4

2 回答 2

2

您收到该错误是因为在函数内部创建的变量超出了该函数的范围，即在评估后续函数时，R不知道该函数的存储值result1和result2在该函数外部使用

function(column)
    {
        result1 <- cor(ref[[column]],sour2[[column]],use = "na.or.complete")
        # calculate using ref and sour1

        result2 <-  cor(ref[[column]],sour1[[column]],use = "na.or.complete")   
        # calculate using ref and sour2

    }

为了演示我刚才所说的，试试下面的代码

testFunc <- function() {
    a <- 0
}

testFunc()

print(a)

你得到如下错误：

Error in print(a) : object 'a' not found

因此，正如 建议的那样，SimonO101您必须单独分配lapply调用结果。sour1sour2

result1 <- lapply( columns , function(column) { cor(ref[[column]],sour2[[column]],use = "na.or.complete") } )

result2 <- lapply( columns , function( column ) { cor(ref[[column]],sour1[[column]],use = "na.or.complete") } )
于 2013-03-15T09:30:32.213 回答
1

您需要将 result1 分配给 lapply.ie

result1 <- lapply( columns , function(column) { cor(ref[[column]],sour2[[column]],use = "na.or.complete") } )

并对结果 2 执行相同操作：

result2 <- lapply( columns , function( column ) { cor(ref[[column]],sour1[[column]],use = "na.or.complete") } )

要分离出您的结果，请尝试更改您的代码以像这样格式化您的结果......

    columns <- paste0("X", 1:13)
result1 <- lapply( columns , function(column){ cor(ref[[column]],sour2[[column]],use = "na.or.complete") } )
result2 <- lapply( columns , function(column){ cor(ref[[column]],sour1[[column]],use = "na.or.complete") } )
names( result1 ) <- columns
names( result2 ) <- columns


write.table( result1 , file = "foo.txt" , row.names = "res1" )
write.table( result2 , file = "foo.txt" , row.names = "res2" , col.names = FALSE  , append = TRUE )
于 2013-03-15T08:59:24.810 回答