如何在r studio中的多个excel工作表上使用脚本

我有一个相当长的剧本，至少对我来说是这样，但我是R的新手

看起来像这样

#DATA IS LOADED
library(xlsx)
df=read.xlsx2("/Users/emiliemariafalkkallenbach/Documents/Norsk_Institut_for_vandforskning/Chase_DK.xlsx",sheetIndex=1)
df[df==""]<- NA 
df=data.matrix(df)
df=data.frame(df)
attach(df)
Phyto=df[,1:6]
Submerged=df[,7:12]
Benthic=df[13:18]


# PHTYOPLANKTON 
#making of boundaries 
a<-(ifelse(P_Resp<=0,(1-P_Acdev),1/(1+P_Acdev)))
b=(0.95+a)/2
c=2*a-b
d=2*c-a
#Weighted boundaries 
e=0.95*P_weight
e=sum(e)
f=b*P_weight
f=sum(f)
g=a*P_weight
g=sum(g)
h=c*P_weight
g=sum(g)
i=d*P_weight
i=sum(i)
#EQR weighted 
j=P_EQR*P_weight
j=sum(j)
#complete
l <- ifelse(is.na(rowSums(Phyto)),1,0)
m=sum(l)


#SUBMERGED VEGETATION
#Overall assesment 
y=ifelse(m<=0,j,"")
z=ifelse(y!="NA",0,(ifelse(y>f,1,(ifelse(y>g,2,(ifelse(y>h,3,(ifelse(y>i,4,5)))))))))


#making of boundaries 
a.sub<-(ifelse(S_Resp<=0,(1-S_Acdev),1/(1+S_Acdev)))
a.sub=sum(a.sub)
b.sub=(0.95+a.sub)/2
b.sub=sum(b.sub)
c.sub=2*a.sub-b.sub
c.sub=sum(c.sub)
d.sub=2*c.sub-a.sub
d.sub=sum(d.sub)
#Weighted boundaries 
e.sub=0.95*S_weight
e.sub=sum(e)
f.sub=b.sub*S_weight
f.sub=sum(f)
g.sub=a.sub*S_weight
g.sub=sum(g)
h.sub=c.sub*S_weight
h.sub=sum(h)
i.sub=d.sub*S_weight
i.sub=sum(i)
#EQR.sub weighted 
j.sub=S_EQR*S_weight
j.sub=sum(j.sub)
#complete.sub
l.sub <- ifelse(is.na(rowSums(Submerged)),1,0)
m.sub=sum(l.sub)
#Overall Assesment
q.sub=m.sub*0.75
y.sub=ifelse(m.sub<=0,j.sub,"")
z.sub=ifelse(y.sub!="NA",(ifelse(y.sub>f.sub,1,(ifelse(y.sub>g.sub,2,(ifelse(y.sub>h.sub,3,(ifelse(y.sub>i.sub,4,5)))))))),0)

BENTHIC INVERTEBRATES
#making of boundaries 
a.ben<-(ifelse(B_Resp<=0,(1-B_Acdev),1/(1+B_Acdev)))
b.ben=(0.95+a.ben)/2
c.ben=2*a.ben-b.ben
d.ben=2*c.ben-a.ben
#Weighted boundaries 
e.ben=0.95*B_weight
e.ben=sum(e.ben)
f.ben=b.ben*B_weight
f.ben=sum(f.ben)
g.ben=a.ben*B_weight
g.ben=(sum(g.ben))
h.ben=c.ben*B_weight
h.ben=sum(h.ben)
i.ben=d.ben*B_weight
i.ben=sum(i.ben)
#EQR weighted 
j.ben=B_EQR*B_weight
#Complete
l.ben <- ifelse(is.na(rowSums(Benthic)),1,0)
m.ben=sum(l.sub)
#ChkAccDev
n.ben=ifelse(B_Resp>0,0.53,1.1)
o.ben=ifelse(B_Acdev<0.15,-1,0)
p.ben=ifelse(B_Acdev>n.ben,1,o.ben)
#Overall Assesment
q.ben=m.ben*0.75
y.ben=ifelse(m.ben<1,j.ben,"")
z.ben=ifelse(y.ben!="NA",ifelse(y.ben>f.ben,1,(ifelse(y.ben>g.ben,2,(ifelse(y.ben>h.ben,3,(ifelse(y.ben>i.ben,4,5))))))),0)

#Final assesment
Z=max(na.omit(c(z,z.sub,z.ben)))

#Overall assesment 
SCORE=(ifelse(Z<=1 && Z<2,"HIGH",(ifelse(Z>=2 && Z<3,"GOOD",(ifelse(Z>=3 && Z<4,"MODERATE",(ifelse(Z>=4 && Z<5,"BAD",(ifelse(Z>=5,"POOR","NA"))))))))))
SCORE


#Pie Chart 
library('plotrix')
Phyto=33.3
Submerged=33.3
Benthic=33.3
total=100
Slices1=c=(1)
total=100
iniR=0.2 
Slices1=c=(1)
pie(1, radius=iniR, init.angle=90, col=c('white'), border = NA, labels='Overall')
colors=c(ifelse(z<=1,"blue", ifelse(z<=2,"green",ifelse(z<=3,"yellow",ifelse(z<=4,"orange",ifelse(z<=5,"red","red"))))), ifelse(z.sub<=1,"blue", ifelse(z.sub<=2,"green", ifelse(z.sub<=3,"yellow",ifelse(z.sub<=4,"orange",ifelse(z.sub<=5,"red","red"))))),ifelse(z.ben<=1,"blue", ifelse(z.ben<=2,"green", ifelse(z.ben<=3, "yellow", ifelse(z.ben<=4,"orange",ifelse(z.ben<=5,"red","red"))))))
floating.pie(0,0,c(Phyto, Submerged, Benthic),radius=5*iniR, startpos=pi/2, col=colors,border=NA)
Slices1=c=(1)
total=100
iniR=0.2 # initial radius
colorst=c(ifelse(Z<=1,"blue", ifelse(Z<=2,"green",ifelse(Z<=3,"yellow",ifelse(Z<=4,"orange",ifelse(Z<=5,"red","red"))))))
floating.pie(0,0,c(total),radius=3*iniR, startpos=pi/2, col=colorst,border=NA)
angles=as.numeric(c(-10,75,80))
pie.labels(0,0,angles,c("Phyto","Submerged","Benthic"),radius=1, bg="white")
pie.labels(0,0,0,c("Overall Assessment"),radius=-0.3)

             z      z.sub     z.ben    Z       Pie-chart (only if possible)

Sheet 1      0        NA        NA    High
Sheet 2     ...       ...       ...   ...
Sheet 3     ...       ...       ...   ... 

如果这是一个普通的问题，我很抱歉

希望有人能帮忙

---

谢谢

将脚本主体包装到for循环中。你有好几次这样做，这里有两次

# pre-allocate an object to write to
out <- matrix(rep(NA, numsheets * numcols), ncols = numcols))
for (i in 1:nsheets) {
  df=read.xlsx2(".../Chase_DK.xlsx", sheetIndex = i)
  #... do calculations
  out[, i] <- c(z, z.sub, z.ben, Z)

}

# second way, no need to pre-allocate anything
sapply(1:nsheets, FUN = function(x) {
  df=read.xlsx2(".../Chase_DK.xlsx", sheetIndex = x)
  #... do calculations
  out <- c(z, z.sub, z.ben, Z) # specify what you wish the function to return
  return(out) # sapply will try to simplify the combined result on its own
})

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阅读xls的更好方法是使用图书馆

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谢谢你的回复。因为某种原因它不起作用。我使用了您建议的第二个解决方案，但在'y=ifelsemIf之后出现了一个错误。如果使用您的第一个建议，则在第一行之后出现了一个错误，表示存在太多的问题。如果我删除其中一个，它会在matrixrepNA中显示错误，28*18，ncols=numcols:unused参数ncols=numcols@EmilieK请投入时间制作一个可复制的示例。这样我们就能为你提供你想要的准确行为。例如，我用一个可复制的例子问了一个问题，在10秒内得到了一个答案。然后写dir并检查是否看到脚本和xls文件。

# remove "#" if you don't have these libraries installed already
# install.packages("readxl")  # faster excel reader
# install.packages("data.table")  # faster everything, in this case rbindlist

library(readxl)
library(data.table)

sheets = 1:5  # index numbers or names of the sheets you want to read

readmysheets = function(sheets) {
  df = read_excel(file="myexcel.xls", sheets)
}

myfiles = lapply(sheets, readmysheets)  # apply the indices/names on the readmysheets function
# you now have a list:
str(myfiles)

# bind the separate sheets together
together = rbindlist(myfiles, fill = T)