R编程帮助编辑代码

问题描述

我已经问了很多问题，所有答案都非常有帮助...但是我的数据又很奇怪，我需要帮助...基本上，我想做的就是找到某个特定速度下的平均速度间隔范围...让我们说从6 s到40 s的平均速度将是5 m/s ...等等.因此有人指出我要使用此代码...

I've asked many questions about this and all the answers were really helpful...but once again my data is weird and I need help...Basically, what I want to do is find the average speed at a certain range of intervals...lets say from 6 s to 40 s my average speed would be 5 m/s...etc etc..So it was pointed out to me to use this code...

library(IRanges)
idx <- seq(1, ncol(data), by=2)
# idx is now 1, 3, 5. It will be passed one value at a time to `i`.
# that is, `i` will take values 1 first, then 3 and then 5 and each time
# the code within is executed.
o <- lapply(idx, function(i) {
    ir1 <- IRanges(start=seq(0, max(data[[i]]), by=401), width=401)
    ir2 <- IRanges(start=data[[i]], width=1)
    t <- findOverlaps(ir1, ir2)
    d <- data.frame(mean=tapply(data[[i+1]], queryHits(t), mean))
    cbind(as.data.frame(ir1), d)
})

给出此输出

# > o
# [[1]]
#   start end width mean
# 1     0 400   401 1.05
#
# [[2]]
#   start end width mean
# 1     0 400   401  1.1
#
# [[3]]
#   start end width     mean
# 1     0 400   401 1.383333

因此，如果我希望它是每100 s ...我只需将 ir1<-.....，by = 401 更改为 by = 100 .

So if I wanted it to be every 100 s... I'll just change ir1 <- ....., by = 401 to become by=100.

但是由于一些原因，我的数据很奇怪

But my data is weird because of a few things

1. 我的数据并非总是以0 s开头，有时是20 s ...根据标本及其是否移动而定
2. 我的数据收集不是每1s或2s或3s进行一次.因此，有时我会得到1-20 s的数据，但是仅仅因为标本不动而跳过了20-40 s.
3. 我认为代码的 findOverlaps 部分会影响我的输出.如何在不干扰输出的情况下摆脱这种情况?

1. my data doesnt always start with 0 s sometimes it starts at 20 s...depending on the specimen and whether it moves
2. My data collection does not happen every 1s or 2s or 3s. Hence sometimes I get data 1-20 s but it skips over 20-40 s simply because the specimen does not move.
3. I think the findOverlaps portion of the code affects my output. How can I get rid of that without disturbing the output?

这里有一些数据可以说明我的麻烦...但是我所有的真实数据都在2000年代结束

Here is some data to illustrate my troubles...but all of my real data ends in 2000s

Time    Speed   Time    Speed   Time    Speed
6.3 1.6 3.1 1.7 0.3 2.4
11.3    1.3 5.1 2.2 1.3 1.3
13.8    1.3 6.3 3.4 3.1 1.5
14.1    1.0 7.0 2.3 4.5 2.7
47.4    2.9 11.3    1.2 5.1 0.5
49.2    0.7 26.5    3.3 5.9 1.7
50.5    0.9 27.3    3.4 9.7 2.4
57.1    1.3 36.6    2.5 11.8    1.3
72.9    2.9 40.3    1.1 13.1    1.0
86.6    2.4 44.3    3.2 13.8    0.6
88.5    3.4 50.9    2.6 14.0    2.4
89.0    3.0 62.6    1.5 14.8    2.2
94.8    2.9 66.8    0.5 15.5    2.6
117.4   0.5 67.3    1.1 16.4    3.2
123.7   3.2 67.7    0.6 26.5    0.9
124.5   1.0 68.2    3.2 44.7    3.0
126.1   2.8 72.1    2.2 45.1    0.8

从数据中可以看到，它不一定要在60 s内结束，有时有时只能在57 s上结束

As you can see from the data, it doesnt necessarily end in 60 s etc sometimes it only ends at 57 etc

编辑添加数据处理

structure(list(Time = c(6.3, 11.3, 13.8, 14.1, 47.4, 49.2, 50.5,
57.1, 72.9, 86.6, 88.5, 89, 94.8, 117.4, 123.7, 124.5, 126.1),
    Speed = c(1.6, 1.3, 1.3, 1, 2.9, 0.7, 0.9, 1.3, 2.9, 2.4,
    3.4, 3, 2.9, 0.5, 3.2, 1, 2.8), Time.1 = c(3.1, 5.1, 6.3,
    7, 11.3, 26.5, 27.3, 36.6, 40.3, 44.3, 50.9, 62.6, 66.8,
    67.3, 67.7, 68.2, 72.1), Speed.1 = c(1.7, 2.2, 3.4, 2.3,
    1.2, 3.3, 3.4, 2.5, 1.1, 3.2, 2.6, 1.5, 0.5, 1.1, 0.6, 3.2,
    2.2), Time.2 = c(0.3, 1.3, 3.1, 4.5, 5.1, 5.9, 9.7, 11.8,
    13.1, 13.8, 14, 14.8, 15.5, 16.4, 26.5, 44.7, 45.1), Speed.2 = c(2.4,
    1.3, 1.5, 2.7, 0.5, 1.7, 2.4, 1.3, 1, 0.6, 2.4, 2.2, 2.6,
    3.2, 0.9, 3, 0.8)), .Names = c("Time", "Speed", "Time.1",
"Speed.1", "Time.2", "Speed.2"), class = "data.frame", row.names = c(NA,
-17L))

推荐答案

对不起，如果我不能完全理解您的问题，您能否解释一下为什么这个示例没有按照您的意愿去做?

sorry if i don't understand your question entirely, could you explain why this example doesn't do what you're trying to do?

# use a pre-loaded data set
mtcars

# choose which variable to cut
var <- 'mpg'

# define groups, whether that be time or something else
# and choose how to cut it.
x <- cut( mtcars[ , var ] , c( -Inf , seq( 15 , 25 , by = 2.5 ) , Inf ) )

# look at your cut points, for every record
x

# you can merge them back on to the mtcars data frame if you like..
mtcars$cutpoints <- x
# ..but that's not necessary

# find the mean within those groups
tapply(
    mtcars[ , var ] ,
    x ,
    mean
)


# find the mean within groups, using a different variable
tapply(
    mtcars[ , 'wt' ] ,
    x ,
    mean
)

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