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我正在使用移动平均线和分箱制作两个直方图。通过使用 excel,我得到了 18k 数据点的移动平均值,其中大多数是 0 值。

这就是我希望通过 R 完成的事情

“移动平均线”

我想使用 R 来制作一个脚本,该脚本将生成一个设备收到多少“计数”的直方图。我努力了 :

hist(y, 20)  
hist (y, ) 
plot (y, x) 

现在,经过三天的学习,这就是我得到的:

y <- AltWithAllCounts$Cts.p.ms
x <- AltWithAllCounts$Alt 
barwidth <- 100 
#how many bins
block <- rep(seq(1,length(x)/barwidth),each=barwidth)
#makes bins
a <- aggregate(y,by=list(block),sum) 
#creates sum of bins
altmean <- aggregate(x,by=list(block),mean)
#finds mean altitude of each bin
avgCount <- a$x/barwidth
#averages out each bin
plot(altmean$x,avgCount,xlab="Altitude",ylab="Counts") 
# creates scatterplot of mean bins
 avgBinCnt <- data.frame(altmean$x,a$x)
write.csv(avgBinCnt,file="avgBinCnt.csv",)

这个想法是我想要 20 个值的平均总和并随着时间的推移绘制它,即 x

x       y
851304  0
851404  0
851503  0
851603  1
851703  0
851804  0
851904  0
852107  0
852203  0
852303  0
922503  0
922603  2
922703  0
922804  0
922904  0
923107  0
923203  0
923303  0
923404  0
923504  0
923604  0
923703  0
923803  0
923904  0
924108  0
924205  1
1441603 0
1441703 0
1441804 0
1441904 0
1442107 1
1442203 1
1442304 0
1442404 4
1442504 0
1442605 1
1442703 6
1442803 8
1442904 0 
4

1 回答 1

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直方图显示频率,而不是间隔中的出现次数。要获得后者,可以执行以下操作:

# First create some test data
t <- seq(1,20000)
p <- 2000
s <- (sin(t*pi/p)+1)/2
d <- ifelse(runif(length(s))<s,1,0)
# Each element of d now contains a 1 or a 0, with a probability that varies
# according to the sign function
# Choose how many elements to count over
barwidth <- 100
# Create a vector of block numbers, with each numbered block having a length of 
# barwidth
block <- rep(seq(1,length(s)/barwidth),each=barwidth)
# Now we aggregate with the sum to find the number of 1s in each block
a <- aggregate(d,by=list(block),sum)
# And plot it to show that we have the expected result
barplot(a$x)

... 这使:

在此处输入图像描述

对于频率散点图而不是计数条形图,这给出了所需的输出:

midpoint <- aggregate(t,by=list(block),mean)
plot(midpoint$x,a$x,xlab="",ylab="frequency")

或者可以找到并绘制对称的运行平均值:

filt <- rep(1/barwidth,barwidth)
y_sym <- filter(d, filt, sides=2)
plot(t,y_sym,xlab="",ylab="frequency")
于 2014-08-10T00:59:40.857 回答