我通过使用移动平均线和分级来制作两个直方图。通过使用excel,我得到了18k数据点的移动平均值,大多数是0值。
这是我希望通过R
完成的我想使用R来制作一个脚本,该脚本将生成一个设备接收的“计数”的直方图。我试过了:
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",)
我的想法是,我想要20个值的平均总和并随时间绘制它,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
答案 0 :(得分:0)
直方图显示频率,而不是间隔中出现的次数。为了获得后者,可以做这样的事情:
# 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$x)
...给出:
对于频率的散点图而不是计数条形图,这会得到所需的输出:
midpoint <- aggregate(t,by=list(block),mean)
plot(midpoint$x,a$x,xlab="",ylab="frequency")
或者可以找到对称的运行平均值并用:
绘制filt <- rep(1/barwidth,barwidth)
y_sym <- filter(d, filt, sides=2)
plot(t,y_sym,xlab="",ylab="frequency")