Question

我目前正在使用包含数千条记录的数据文件，我必须使用R。

格式化它们

这就是我的数据框现在的样子

    ROWID  CAT SERIALNO SUB PRODUCTNAME HOMENUM Start.X Start.Y End.X End.Y
1 111111111 CATA       10  43  PRODUCT A1      NA      NA      NA    NA    NA
2         1            NA  NA                  NA      NA      NA    NA    NA
3         2    3       NA  NA                  NA      NA      NA    NA    NA
4         4    5       NA  NA                  NA      NA      NA    NA    NA
5 555555555 CATB       13  76  PRODUCT A2      NA      NA      NA    NA    NA
6         6            NA  NA                  NA      NA      NA    NA    NA
7         7    8       NA  NA                  NA      NA      NA    NA    NA
8         9   10       NA  NA                  NA      NA      NA    NA    NA

我想要的格式

    ROWID  CAT SERIALNO SUB PRODUCTNAME HOMENUM Start.X Start.Y End.X End.Y
1 111111111 CATA       10  43  PRODUCT A1       1       2       3     4     5
2 555555555 CATB       13  76  PRODUCT A2       6       7       8     9    10

从上面的第一个截图中可以看出，最后4列的值分别为2,3,4和6,7,8行

我尝试使用t（）函数，但似乎没有产生我需要的东西，并且使用fix（）函数手动排列数据是不可能的，因为我正在处理大型数据文件。

无论如何使用R？

来实现所需的格式

编辑：dput（）的结果

structure(list(V1 = structure(c(9L, 2L, 1L, 3L, 4L, 5L, 6L, 7L, 
8L), .Label = c("1", "111111111", "2", "4", "555555555", "6", 
"7", "9", "ROWID"), class = "factor"), V2 = structure(c(6L, 7L, 
1L, 3L, 4L, 8L, 1L, 5L, 2L), .Label = c("", "10", "3", "5", "8", 
"CAT", "CATA", "CATB"), class = "factor"), V3 = structure(c(4L, 
2L, 1L, 1L, 1L, 3L, 1L, 1L, 1L), .Label = c("", "10", "13", "SERIALNO"
), class = "factor"), V4 = structure(c(4L, 2L, 1L, 1L, 1L, 3L, 
1L, 1L, 1L), .Label = c("", "43", "76", "SUB"), class = "factor"), 
    V5 = structure(c(4L, 2L, 1L, 1L, 1L, 3L, 1L, 1L, 1L), .Label = c("", 
    "PRODUCT A1", "PRODUCT A2", "PRODUCTNAME"), class = "factor"), 
    V6 = structure(c(2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", 
    "HOMENUM"), class = "factor"), V7 = structure(c(2L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", "Start X"), class = "factor"), 
    V8 = structure(c(2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", 
    "Start Y"), class = "factor"), V9 = structure(c(2L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", "End X"), class = "factor"), 
    V10 = structure(c(2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", 
    "End Y"), class = "factor")), .Names = c("V1", "V2", "V3", 
"V4", "V5", "V6", "V7", "V8", "V9", "V10"), class = "data.frame", row.names = c(NA, 
-9L))

Answer 1

我怀疑你应该先处理数据导入。您的导入未使用header = TRUE，因此我们必须先修复列名称：

names(DF) <- as.character(unlist(DF[1,]))
DF <- DF[-1,]

然后我们可以选择每第五行：

DF1 <- DF[seq_len(nrow(DF)) %% 4 == 1L,]

现在我们可以选择所有其他行的前两列并转置它们：

temp <- t(DF[seq_len(nrow(DF)) %% 4 != 1L, 1:2])

我们从生成的字符矩阵中删除空单元格，并将生成的字符向量转换为五列矩阵，然后将其分配给新data.frame的最后五列：

DF1[, 6:10] <- matrix(temp[temp != ""], ncol = 5, byrow = TRUE)

最后，我们转换列类型，因此该数字实际上是数字而不是字符：

DF1[] <- lapply(DF1, function(x) type.convert(as.character(x), as.is = TRUE))
print(DF1)
#      ROWID  CAT SERIALNO SUB PRODUCTNAME HOMENUM Start X Start Y End X End Y
#2 111111111 CATA       10  43  PRODUCT A1       1       2       3     4     5
#6 555555555 CATB       13  76  PRODUCT A2       6       7       8     9    10

R中的数据格式？

1 个答案: