Question

我有一个包含两个if语句的for循环。一切正常，除了我想将第二个if语句传递给变量v1的值。我首先分配变量，变量充满了很多NA和我期待的三个值。但我只想在变量中使用这三个值。

因此，不是只编写满足要求的i，而是将每个i写入变量。

v1 <- c()

for(i in seq_along(y$values)){
  if(y$lengths[i] < 500 && y$values[i] == TRUE){
    y$values[i] <- FALSE
  }
  if(y$lengths[i] > 500 && y$values[i] == TRUE){
    print(y$lengths[i])
    v1[[i]] <- y$lengths[i]
  }
}

我查看了apply函数，但无法真正弄清楚如何在apply中使用这两个if语句。

数据是一个很大的逻辑向量。我在该向量上使用y <- rle(vec)，它给了y$lengths（1到几十万之间的所有内容）和y$values（TRUE和FALSE）。

所需的结果是rle长度> 500且TRUE为真。：

[1] 5120
[1] 16630
[1] 10188

哪个打印命令让我很好。

dput(y$length)给了我：

c(129719L, 1L, 79337L, 2L, 4L, 1L, 3L, 1L, 2L, 1L, 1L, 1L, 4L, 
2L, 2L, 3L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 5L, 1L, 3L, 1L, 
6L, 1L, 5120L, 1L, 39L, 1L, 12L, 1L, 121L, 1L, 14L, 1L, 6L, 1L, 
3L, 1L, 3L, 1L, 2L, 1L, 6L, 1L, 11L, 1L, 9L, 1L, 10L, 2L, 6L, 
1L, 2L, 2L, 1L, 1L, 7L, 2L, 4L, 1L, 2L, 1L, 4L, 1L, 3L, 2L, 5L, 
1L, 5L, 4L, 8L, 1L, 4L, 1L, 4L, 2L, 2L, 2L, 9L, 1L, 2L, 1L, 2L, 
1L, 3L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 1L, 3L, 3L, 1L, 
7L, 1L, 1L, 1L, 1L, 2L, 6L, 2L, 2L, 1L, 2L, 4L, 3L, 1L, 1L, 1L, 
4L, 3L, 2L, 1L, 5L, 5L, 2L, 2L, 3L, 1L, 2L, 1L, 1L, 1L, 2L, 1L, 
4L, 2L, 2L, 1L, 1L, 1L, 4L, 1L, 1L, 3L, 4L, 2L, 1L, 1L, 13L, 
1L, 3L, 2L, 3L, 1L, 9L, 1L, 1L, 1L, 1L, 2L, 3L, 1L, 4L, 1L, 13L, 
1L, 3L, 1L, 4L, 1L, 8L, 1L, 7L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 3L, 
1L, 4L, 1L, 2L, 2L, 5L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 1L, 4L, 3L, 
3L, 2L, 1L, 1L, 2L, 1L, 1L, 2L, 3L, 1L, 2L, 1L, 4L, 1L, 9L, 2L, 
6L, 1L, 14L, 1L, 2L, 1L, 6L, 1L, 16630L, 7L, 1L, 3L, 2L, 3L, 
1L, 4L, 3L, 4L, 1L, 1L, 2L, 7L, 1L, 1L, 1L, 1L, 1L, 8L, 1L, 5L, 
1L, 6L, 1L, 1L, 3L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 2L, 
1L, 1L, 1L, 6L, 1L, 5L, 1L, 2L, 2L, 3L, 1L, 5L, 2L, 3L, 1L, 2L, 
2L, 10L, 1L, 4L, 1L, 2L, 2L, 1L, 2L, 1L, 1L, 1L, 1L, 7L, 3L, 
1L, 1L, 4L, 1L, 1L, 1L, 10L, 1L, 2L, 2L, 3L, 1L, 3L, 5L, 2L, 
3L, 1L, 1L, 2L, 1L, 1L, 3L, 1L, 4L, 3L, 1L, 1L, 1L, 1L, 2L, 2L, 
1L, 4L, 2L, 1L, 1L, 3L, 1L, 1L, 2L, 1L, 4L, 1L, 1L, 3L, 4L, 2L, 
1L, 2L, 1L, 1L, 3L, 1L, 2L, 1L, 2L, 1L, 3L, 1L, 2L, 1L, 3L, 1L, 
1L, 1L, 3L, 2L, 12L, 1L, 3L, 2L, 2L, 1L, 1L, 2L, 7L, 1L, 2L, 
1L, 1L, 2L, 1L, 1L, 7L, 1L, 2L, 1L, 4L, 1L, 7L, 1L, 4L, 1L, 1L, 
1L, 6L, 1L, 6L, 1L, 6L, 2L, 14L, 1L, 5L, 1L, 9L, 1L, 1L, 1L, 
1L, 2L, 39L, 1L, 20L, 1L, 1L, 1L, 6L, 1L, 9L, 2L, 5L, 1L, 7L, 
1L, 16L, 1L, 22L, 1L, 1L, 1L, 10L, 1L, 20L, 1L, 18L, 1L, 20L, 
1L, 3L, 1L, 2L, 1L, 2L, 1L, 5L, 1L, 9L, 1L, 3L, 1L, 3L, 1L, 15L, 
1L, 10L, 1L, 40L, 1L, 30L, 1L, 111L, 1L, 314L, 1L, 9L, 1L, 10188L, 
4L, 88L, 1L, 8L, 1L, 1L, 1L, 1L, 1L, 15L, 1L, 24L, 1L, 1L, 2L, 
3L, 1L, 16L, 1L, 3L, 1L, 4L, 2L, 2L, 2L, 1L, 2L, 5L, 2L, 2L, 
7L, 1L, 1L, 3L, 2L, 3L, 7L, 2L, 1L, 1L, 6L, 1L, 4L, 2L, 2L, 1L, 
2L, 1L, 1L, 1L, 13L, 1L, 2L, 1L, 2L, 1L, 3L, 1L, 3L, 1L, 4L, 
1L, 16L, 1L, 4071L, 5L, 162912L, 1L, 6L, 1L, 280986L)

Answer 1

你可以非常有效地获得最终的v1向量而不用循环或像这样应用：

v1 <- y$lengths[y$lengths > 500 & y$values == TRUE]

如果你想使用循环，这里有两个选项：

（1）您可以像这样附加到v1：

v1 <- c()
for(i in seq_along(y$values)){
  if(y$lengths[i] < 500 && y$values[i] == TRUE){
    y$values[i] <- FALSE
  }
  if(y$lengths[i] > 500 && y$values[i] == TRUE){
    print(y$lengths[i])
    v1 <- c(v1, y$lengths[i])
  }
}

（2）如果向量v1真的很长，那么像这样的追加可能会变慢。作为替代方案，您可以预先分配，然后最终排除这些NAs：

v1 <- rep(NA, length)
for(i in seq_along(y$values)){
  if(y$lengths[i] < 500 && y$values[i] == TRUE){
    y$values[i] <- FALSE
  }
  if(y$lengths[i] > 500 && y$values[i] == TRUE){
    print(y$lengths[i])
    v1[i] <- y$lengths[i]
  }
}
v1 <- v1[!is.na(v1)]

Answer 2

这似乎适用于我通过阅读您的问题推断出的样本数据：

error

**

y<-data.frame(lengths=seq(100,900,100), 
              values=c("TRUE", "FALSE", "TRUE", "FALSE", "TRUE", "FALSE", "TRUE", "FALSE", "TRUE"))
v1 <- c()

for(i in seq_along(y$values)){
  if(y$lengths[i] < 500 && y$values[i] == TRUE){
y$values[i] <- FALSE
  }
  if(y$lengths[i] > 500 && y$values[i] == TRUE){
#print(y$lengths[i])
ifelse(length(v1)>0, v1<-rbind(v1, y$lengths[i]), v1<-y$lengths[i])
#v1 <- rbind(y$lengths[i])
  }
}

**

R - 在两个语句之后将值写入变量

2 个答案: