Question

我有一个数据框foo，其中包含4个空列，我需要填充列表result的内容。我不能通过公共ID进行合并，因为其中一个不存在，但foo和result长度相同。如何访问result的每个元素的内容并将其转换为数据帧foo的列？

我理解要访问单个列表元素的每个部分，必须写：result[[1]][1]，result[[1]][2]，result[[1]][3]，result[[1]][4]，但是如何迭代他们？

我试过了

for (i in 1:length(result)){

   foo$coords.x1.min <- result[[i]][1]
}

但这仅使用列表中的第一个值填充整个列

> nrow(foo)
[1] 103
> length(result)
[1] 103

 > head(foo)
     names            X1            X2            X3            X4
1  4CITIES            NA            NA            NA            NA
2     ACES            NA            NA            NA            NA
3 ADVANCES            NA            NA            NA            NA
4    AMASE            NA            NA            NA            NA
5  ARCHMAT            NA            NA            NA            NA
6      ASC            NA            NA            NA            NA


> head(result)
[[1]]
              min     max
coords.x1 -3.7038 16.3738
coords.x2  5.6761 50.8503

[[2]]
              min     max
coords.x1 -5.4710 25.1442
coords.x2 35.3387 56.4152

[[3]]
              min     max
coords.x1 -9.1393 21.0122
coords.x2 38.7223 57.0488

[[4]]
              min     max
coords.x1  2.1734 22.1567
coords.x2 41.3851 65.5848

[[5]]
              min     max
coords.x1 -7.9093 22.9444
coords.x2 38.5707 41.9028

[[6]]
              min     max
coords.x1  3.0573 24.9384
coords.x2 44.4949 60.1699

Answer 1

这是一个解决方案

 for (i in 1:length(result)){

    foo$X1[i] <- result[[i]][1]
    foo$X2[i] <- result[[i]][2]
    foo$X3[i] <- result[[i]][3]
    foo$X4[i] <- result[[i]][4]
 }

Answer 2

我们可以通过unlist'结果'在没有循环的情况下执行此操作，将其转换为包含4列的matrix，并将其分配给'foo'列，但第一列除外一个

foo[-1] <- matrix(unlist(result), ncol=4, byrow = TRUE)
foo
#     names      X1      X2      X3      X4
#1  4CITIES -3.7038  5.6761 16.3738 50.8503
#2     ACES -5.4710 35.3387 25.1442 56.4152
#3 ADVANCES -9.1393 38.7223 21.0122 57.0488
#4    AMASE  2.1734 41.3851 22.1567 65.5848
#5  ARCHMAT -7.9093 38.5707 22.9444 41.9028
#6      ASC  3.0573 44.4949 24.9384 60.1699

数据

result <- list(structure(c(-3.7038, 5.6761, 16.3738, 50.8503), .Dim = c(2L, 
2L), .Dimnames = list(c("coords.x1", "coords.x2"), c("min", "max"
))), structure(c(-5.471, 35.3387, 25.1442, 56.4152), .Dim = c(2L, 
2L), .Dimnames = list(c("coords.x1", "coords.x2"), c("min", "max"
))), structure(c(-9.1393, 38.7223, 21.0122, 57.0488), .Dim = c(2L, 
2L), .Dimnames = list(c("coords.x1", "coords.x2"), c("min", "max"
))), structure(c(2.1734, 41.3851, 22.1567, 65.5848), .Dim = c(2L, 
2L), .Dimnames = list(c("coords.x1", "coords.x2"), c("min", "max"
))), structure(c(-7.9093, 38.5707, 22.9444, 41.9028), .Dim = c(2L, 
2L), .Dimnames = list(c("coords.x1", "coords.x2"), c("min", "max"
))), structure(c(3.0573, 44.4949, 24.9384, 60.1699), .Dim = c(2L, 
2L), .Dimnames = list(c("coords.x1", "coords.x2"), c("min", "max"
))))


foo <- structure(list(names = c("4CITIES", "ACES", "ADVANCES", "AMASE", 
"ARCHMAT", "ASC"), X1 = c(NA, NA, NA, NA, NA, NA), X2 = c(NA, 
NA, NA, NA, NA, NA), X3 = c(NA, NA, NA, NA, NA, NA), X4 = c(NA, 
NA, NA, NA, NA, NA)), .Names = c("names", "X1", "X2", "X3", "X4"
), class = "data.frame", row.names = c("1", "2", "3", "4", "5", 
"6"))

填充列表

2 个答案:

数据