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在R之间没有for循环的行之间移动值

[英]Moving values between rows without a for loop in R

I have written some code used to organize data sampled at different frequencies, but I made extensive use of for-loops, which slow the code's operation down significantly when the data set is large. 我编写了一些用于组织以不同频率采样的数据的代码,但是我广泛使用了for循环,当数据集很大时,这会大大降低代码的操作速度。 I've been going through my code, finding ways to remove for-loops to speed it up, but one of the loops has got me stumped. 我一直在浏览我的代码,找到删除for循环以加快速度的方法,但其中一个循环让我感到难过。

As an example, let's say the data was sampled at 3Hz, so I get three rows for every second of data. 举个例子,假设数据是以3Hz采样的,所以每秒数据都会得到三行。 However, the variables A, B, and C are sampled at 1Hz each, so I will get one value every three rows for each of them. 但是,变量A,B和C各自以1Hz采样,因此每三行我将获得一个值。 The variables are sampled consecutively within the one second period, resulting in a diagonal nature to the data. 在一秒周期内连续采样变量,导致数据的对角性质。

To further complicate things, sometimes a row is lost in the original data set. 更复杂的是,有时在原始数据集中会丢失一行。

My goal is this: Having identified the rows that I wish to keep, I want to move the non-NA values from the subsequent rows up into the keeper rows. 我的目标是:确定了我希望保留的行后,我想将后续行中的非NA值移动到守护者行中。 If it weren't for the lost data issue, I would always keep the row containing a value for the first variable, but if one of these rows is lost, I will be keeping the next row. 如果它不是丢失的数据问题,我将始终保持行包含第一个变量的值,但如果其中一行丢失,我将保留下一行。

In the example below, the sixth sample and the tenth sample are lost. 在下面的示例中,第六个样本和第十个样本丢失。

A <- c(1, NA, NA, 4, NA, 7, NA, NA, NA, NA)
B <- c(NA, 2, NA, NA, 5, NA, 8, NA, 11, NA)
C <- c(NA, NA, 3, NA, NA, NA, NA, 9, NA, 12)

test_df <- data.frame(A = A, B = B, C = C)

test_df
     A  B  C
 1   1 NA NA
 2  NA  2 NA
 3  NA NA  3
 4   4 NA NA
 5  NA  5 NA
 6   7 NA NA
 7  NA  8 NA
 8  NA NA  9
 9  NA 11 NA
10  NA NA 12

keep_rows <- c(1, 4, 6, 9)

After I move the values up into the keeper rows, I will remove the interim rows, resulting in the following: 将值移动到守护者行后,我将删除临时行,从而产生以下结果:

test_df <- test_df[keep_rows, ]
test_df
     A  B  C
 1   1  2  3
 2   4  5 NA
 3   7  8  9
 4  NA 11 12

In the end, I only want one row for each second of data, and NA values should only remain where a row of the original data was lost. 最后,我只希望每秒数据有一行,而NA值只应保留原始数据行丢失的位置。

Does anyone have any ideas of how to move the data up without using a for-loop? 有没有人有任何关于如何在不使用for循环的情况下移动数据的想法? I'd appreciate any help! 我很感激任何帮助! Sorry if this question is too wordy; 对不起,如果这个问题太罗嗦了; I wanted to err on the side of too much information rather than not enough. 我想在太多信息方面犯错,而不是不够。

This should do it: 这应该这样做:

test_df = with(test_df, cbind(A[1:(length(A)-2)], B[2:(length(B)-1)], C[3:length(C)]))
test_df = data.frame(test_df[!apply(test_df, 1, function(x) all(is.na(x))), ])
colnames(test_df) = c('A', 'B', 'C')
> test_df
   A  B  C
1  1  2  3
2  4  5 NA
3  7  8  9
4 NA 11 12

And if you want something even faster : 如果你想要更快的东西:

test_df = data.frame(test_df[rowSums(is.na(test_df)) != ncol(test_df), ])

Building on the great answer by @John Colby, we can get rid of the apply step and speed it up quite a bit (about 20x): 在@John Colby的伟大答案的基础上,我们可以摆脱应用步骤并加速相当多(大约20倍):

# Create a bigger test set 
A <- c(1, NA, NA, 4, NA, 7, NA, NA, NA, NA)
B <- c(NA, 2, NA, NA, 5, NA, 8, NA, 11, NA)
C <- c(NA, NA, 3, NA, NA, NA, NA, 9, NA, 12)
n=1e6; test_df = data.frame(A=rep(A, len=n), B=rep(B, len=n), C=rep(C, len=n))

# John Colby's method, 9.66 secs
system.time({
  df1 = with(test_df, cbind(A[1:(length(A)-2)], B[2:(length(B)-1)], C[3:length(C)]))
  df1 = data.frame(df1[!apply(df1, 1, function(x) all(is.na(x))), ])
  colnames(df1) = c('A', 'B', 'C')
})

# My method, 0.48 secs
system.time({
  df2 = with(test_df, data.frame(A=A[1:(length(A)-2)], B=B[2:(length(B)-1)], C=C[3:length(C)]))
  df2 = df2[is.finite(with(df2, A|B|C)),]
  row.names(df2) <- NULL
})

identical(df1, df2) # TRUE

...The trick here is that A|B|C is only NA if all values are NA . ...这里的技巧是, A|B|C是唯一的NA ,如果所有的值都是NA This turns out to be much faster than calling all(is.na(x)) on each row of a matrix using apply . 这比使用apply在矩阵的每一行上调用all(is.na(x))要快得多。

EDIT @John has a different approach that also speeds it up. 编辑 @John有一个不同的方法,也加快了它。 I added some code to turn the result into a data.frame with correct names and timed it. 我添加了一些代码将结果转换为具有正确名称的data.frame并定时。 It seems to be pretty much the same speed as my solution. 它似乎与我的解决方案速度几乎相同。

# John's method, 0.50 secs
system.time({
  test_m = with(test_df, cbind(A[1:(length(A)-2)], B[2:(length(B)-1)], C[3:length(C)]))
  test_m[is.na(test_m)] <- -1
  test_m <- test_m[rowSums(test_m) > -3,]
  test_m[test_m == -1] <- NA
  df3 <- data.frame(test_m)
  colnames(df3) = c('A', 'B', 'C')
})

identical(df1, df3) # TRUE

EDIT AGAIN ...and @John Colby's updated answer is even faster! 再次编辑 ...... @John Colby的更新答案更快!

# John Colby's method, 0.39 secs
system.time({
  df4 = with(test_df, cbind(A[1:(length(A)-2)], B[2:(length(B)-1)], C[3:length(C)]))
  df4 = data.frame(df4[rowSums(is.na(df4)) != ncol(df4), ])
  colnames(df4) = c('A', 'B', 'C')
})

identical(df1, df4) # TRUE

So your question was just about moving up without a loop. 所以你的问题只是在没有循环的情况下向上移动。 So apparently you solved the first step already. 显然你已经解决了第一步。

> test_m <- with( test_df, cbind(A[1:(length(A)-2)], B[2:(length(B)-1)], C[3:length(C)]) )
> test_m
     [,1] [,2] [,3]
[1,]    1    2    3
[2,]   NA   NA   NA
[3,]   NA   NA   NA
[4,]    4    5   NA
[5,]   NA   NA   NA
[6,]    7    8    9
[7,]   NA   NA   NA
[8,]   NA   11   12

Which is now a matrix. 现在是一个矩阵。 You can easily eliminate the rows for which there is no data point now without a loop. 在没有循环的情况下,您可以轻松地消除现在没有数据点的行。 If you want it back to a data.frame then you could use a different method but this one will run the fastest for a large mount of data. 如果您希望它返回到data.frame,那么您可以使用不同的方法,但是这个方法对于大量数据运行速度最快。 I like to just make the NA's an impossible value... perhaps -1 but you'll know best for your data... perhaps -pi. 我想让NA成为一个不可能的价值......也许-1但你知道你的数据最好......也许-pi。

test_m[is.na(test_m)] <- -1

And now just select the rows for a property of those impossible numbers 现在只需为那些不可能数字的属性选择行

test_m <- test_m[rowSums(test_m) > -3,]

And, if you want you can put the NA's back. 并且,如果你想,你可以把NA放回去。

test_m[test_m == -1] <- NA
test_m
     [,1] [,2] [,3]
[1,]    1    2    3
[2,]    4    5   NA
[3,]    7    8    9
[4,]   NA   11   12

There's no loop ( for or apply ) and the one function applied across rows of the matrix is specially optimized and runs very fast (rowSums). 没有循环( forapply ),并且跨矩阵行应用的一个函数经过特殊优化并且运行速度非常快(rowSums)。

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