Question

r-noob这里有一个不能自己解决的问题

我有一个长度为250的数据框

    head(abto, 20)
    See Transekt plant Blatt# breiteo breitez bez
 1: ABT        A     1      1  2.0182  5.3980   1
 2: ABT        A     1      1  1.9730  4.2522   1
 3: ABT        A     1      1  1.8024  3.7587   1
 4: ABT        A     1      2  2.2081  4.2880   2
 5: ABT        A     1      2  2.2858  6.1115   2
 6: ABT        A     1      2  1.8532  5.7426   2
 7: ABT        A     1      3  2.0384  4.9074   2
 8: ABT        A     1      3  2.0757  4.8801   2
 9: ABT        A     1      3  1.8034  4.6111   2
10: ABT        A     1      4  1.9567  4.8879   2
11: ABT        A     1      4  1.9080  5.0652   2
12: ABT        A     1      4  1.8346  4.8862   2
13: ABT        A     1      5  2.0282  4.5545   1
14: ABT        A     1      5  2.1356  5.7157   1
15: ABT        A     1      5  1.7594  6.1688   1
16: ABT        A     2      1  1.6457  5.2868   1
17: ABT        A     2      1  1.6942  5.0414   1
18: ABT        A     2      1  2.0544  5.6711   1
19: ABT        A     2      2  2.1342  5.2867   2
20: ABT        A     2      2  1.9107  6.2139   2

和一个长度为84的向量（只有第三个减去2作为第一个）

> head(databtzl)
  abtl.wert
1   22.7738
2   24.9137
3   24.9474
4   25.0498
5   25.0431
6   21.1024

基本上我有一个数据帧，每个叶子有3个宽度测量值（Blatt＃）和一个测量长度的矢量。我试图将它们合并为一个，但是将长度值放在每第3行。

我想有很多解决方案，我想在向量的每个值之间创建2个空白单元格然后合并它们。

由于数据帧中存在一些不规则性（宽度的两次只有2个而不是3个），每当abto $ Blatt＃的值发生变化时，我就会想到一个函数/循环来分配向量的值。

抱歉这个麻烦的问题，我希望有人理解我的问题。

提前致谢！

编辑：我的预期输出是下表，abtl.wert是databtzl向量中的值......

          See Transekt plant Blatt# breiteo breitez bez  length
     1: ABT        A     1      1  2.0182  5.3980   1  abtl.wert1
     2: ABT        A     1      1  1.9730  4.2522   1
     3: ABT        A     1      1  1.8024  3.7587   1
     4: ABT        A     1      2  2.2081  4.2880   2  abtl.wert2
     5: ABT        A     1      2  2.2858  6.1115   2
     6: ABT        A     1      2  1.8532  5.7426   2
     7: ABT        A     1      3  2.0384  4.9074   2  abtl.wert3
     8: ABT        A     1      3  2.0757  4.8801   2
     9: ABT        A     1      3  1.8034  4.6111   2

Answer 1

如果我理解您的问题，以下内容应该有效：

## Compute the rows in abto to add the values from databtz$abtl.wert
## Here, we look for changes in adjoining rows of the Blatt# column
## using diff. Descending in rows, the row before the change will have
## a diff != 0. We want to mark the next row as well as the first row, 
## so prepend this resulting vector with TRUE
rind <- c(TRUE, diff(abto$`Blatt#`) != 0)

## Then it is just a matter of adding a new column to abto named length
## and inserting the values from databtz1$abtl.wert only for the rows 
## that are marked (identified by which(rind==1))
result <- data.frame(abto, length=rep(NA,nrow(abto)))
result[which(rind==TRUE),"length"] <- databtz1$abtl.wert

此答案假设databtz1中的行数与abto$Blatt#更改的行数相匹配。

使用此输入数据：

abto <- structure(list(See = c("ABT", "ABT", "ABT", "ABT", "ABT", "ABT", 
"ABT", "ABT", "ABT", "ABT", "ABT", "ABT", "ABT", "ABT", "ABT", 
"ABT", "ABT", "ABT", "ABT"), Transekt = c("A", "A", "A", "A", 
"A", "A", "A", "A", "A", "A", "A", "A", "A", "A", "A", "A", "A", 
"A", "A"), plant = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L), `Blatt#` = c(1L, 1L, 1L, 
2L, 2L, 2L, 3L, 3L, 4L, 4L, 4L, 5L, 5L, 5L, 1L, 1L, 1L, 2L, 2L
), breiteo = c(2.0182, 1.973, 1.8024, 2.2081, 2.2858, 1.8532, 
2.0384, 2.0757, 1.9567, 1.908, 1.8346, 2.0282, 2.1356, 1.7594, 
1.6457, 1.6942, 2.0544, 2.1342, 1.9107), breitez = c(5.398, 4.2522, 
3.7587, 4.288, 6.1115, 5.7426, 4.9074, 4.8801, 4.8879, 5.0652, 
4.8862, 4.5545, 5.7157, 6.1688, 5.2868, 5.0414, 5.6711, 5.2867, 
6.2139), bez = c(1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L)), .Names = c("See", "Transekt", 
"plant", "Blatt#", "breiteo", "breitez", "bez"), class = "data.frame", row.names = c(NA, 
-19L))

databtz1 <- structure(list(abtl.wert = c(22.7738, 24.9137, 24.9474, 25.0498, 
25.0431, 21.1024, 28.9083)), .Names = "abtl.wert", class = "data.frame", row.names = c(NA, 
-7L))

我明白了：

print(result)
##       See Transekt plant Blatt. breiteo breitez bez  length
##1  ABT        A     1      1  2.0182  5.3980   1 22.7738
##2  ABT        A     1      1  1.9730  4.2522   1      NA
##3  ABT        A     1      1  1.8024  3.7587   1      NA
##4  ABT        A     1      2  2.2081  4.2880   2 24.9137
##5  ABT        A     1      2  2.2858  6.1115   2      NA
##6  ABT        A     1      2  1.8532  5.7426   2      NA
##7  ABT        A     1      3  2.0384  4.9074   2 24.9474
##8  ABT        A     1      3  2.0757  4.8801   2      NA
##9  ABT        A     1      4  1.9567  4.8879   2 25.0498
##10 ABT        A     1      4  1.9080  5.0652   2      NA
##11 ABT        A     1      4  1.8346  4.8862   2      NA
##12 ABT        A     1      5  2.0282  4.5545   1 25.0431
##13 ABT        A     1      5  2.1356  5.7157   1      NA
##14 ABT        A     1      5  1.7594  6.1688   1      NA
##15 ABT        A     2      1  1.6457  5.2868   1 21.1024
##16 ABT        A     2      1  1.6942  5.0414   1      NA
##17 ABT        A     2      1  2.0544  5.6711   1      NA
##18 ABT        A     2      2  2.1342  5.2867   2 28.9083
##19 ABT        A     2      2  1.9107  6.2139   2      NA

请注意，我从原始发布的数据中删除了Blatt#等于3的行，以模拟您的违规行为，并在databtz1$abtl.wert中添加了一行，以便我们在Blatt# databtz1$abtl.wert行Cells()。

每隔第n行或对应一个规范向数据框添加值。列值

1 个答案: