f = c(1,1)
for(i in 2:30) {
print(sprintf("i=%i, f[i-1]=%i, f[i]=%i, f[i-1]/f[i]=%.4f", i, f[i-1], f[i], f[i]/f[i-1]))
f=c(f, (f[i-1]+f[i])) # sum of last two element in f
} # Q2. 每一步都打印出值,你可以参照一下
> for ( j in 1:15) {
+ print(sprintf("j=%i, answer[j]=%i, 7*answer[j]=%i, 7*answer[j]%%31=%i", j, answer[j], 7*answer[j], (7*answer[j])%%31))
+ answer <- c(answer, ( 7* answer[ j ] ) %% 31)
+ }
[1] "j=1, answer[j]=3, 7*answer[j]=21, 7*answer[j]%31=21"
[1] "j=2, answer[j]=21, 7*answer[j]=147, 7*answer[j]%31=23"
[1] "j=3, answer[j]=23, 7*answer[j]=161, 7*answer[j]%31=6"
[1] "j=4, answer[j]=6, 7*answer[j]=42, 7*answer[j]%31=11"
[1] "j=5, answer[j]=11, 7*answer[j]=77, 7*answer[j]%31=15"
[1] "j=6, answer[j]=15, 7*answer[j]=105, 7*answer[j]%31=12"
[1] "j=7, answer[j]=12, 7*answer[j]=84, 7*answer[j]%31=22"
[1] "j=8, answer[j]=22, 7*answer[j]=154, 7*answer[j]%31=30"
[1] "j=9, answer[j]=30, 7*answer[j]=210, 7*answer[j]%31=24"
[1] "j=10, answer[j]=24, 7*answer[j]=168, 7*answer[j]%31=13"
[1] "j=11, answer[j]=13, 7*answer[j]=91, 7*answer[j]%31=29"
[1] "j=12, answer[j]=29, 7*answer[j]=203, 7*answer[j]%31=17"
[1] "j=13, answer[j]=17, 7*answer[j]=119, 7*answer[j]%31=26"
[1] "j=14, answer[j]=26, 7*answer[j]=182, 7*answer[j]%31=27"
[1] "j=15, answer[j]=27, 7*answer[j]=189, 7*answer[j]%31=3"
>
in 1:nrow(metadata) : 参数长度为零, 可能是metadata内没有数据。
metadata <- jsonlite::fromJSON('metadata.cart.2020-04-22.json')
可能是数据读取失败,检查数据下载是否成功,格式是否正确?
我的建议是可能要大改整个程序:用一个list(比如叫data.list)把所有data frame或者之类的东西装起来,用的时候用data.list[[i]]来取出第i个data frame。用sqldf的时候,第一个和第三个可以用paste来写命令,第二个和第四个可能需要先用一个data frame(比如叫data)来暂时存储data.list里的一个data frame再操作sqldf。总的来说像这样(没怎么用过SQL,可能有问题):
dataold_list<-dataold.list<-datanew_list<-datanew.list<-vector("list",100)for(i in 1:100){
dataold_list[[i]]<-sqldf(paste("select * from Dataold a,num",i," b
where a.LOAN_NO=b.LOAN_NO",sep=""))
dataold_list[[i]]<-dataold_list[[i]][,-1]
data<-dataold_list[[i]]
dataold.list[[i]]<-sqldf("select * from a,data b
where a.LOAN_NO=b.LOAN_NO")
datanew_list[[i]]<-sqldf(paste("select * from Datanew a,num",i," b
where a.LOAN_NO=b.LOAN_NO",sep=""))
datanew_list[[i]]<-dataold_list[[i]][,-1]
data<-datanew_list[[i]]
datanew.list[[i]]<-sqldf("select * from a,data b
where a.LOAN_NO=b.LOAN_NO")
}