load two different excel format at same time-CodePudding

I need to load many .csv and .xlsx(including two or three sheets) in one folder

files <- list.files(path = , pattern = "*.csv")
files <- list.files(path = , pattern = "*.xlsx")

for(i in files) {  }

Can I use loop to load two different excel format at same time? Any ideas? Thank Everyone help!

CodePudding user response：

substring not substr is useful for these kind of vectorised substringing operations:

nc <- seq(nchar(A))
sum(as.numeric(substring(A, nc, nc)) * rev(nc))
#[1] 134

CodePudding user response：

A <- "883745"

First, when given this task, you need to break down the summation and look for pattern. In particular, pay attention to what is changing.

ans  = as.numeric(substr(A , 1, 1)) * (nchar(A) - 0)
ans  = as.numeric(substr(A , 2, 2)) * (nchar(A) - 1)
ans  = as.numeric(substr(A , 3, 3)) * (nchar(A) - 2)
ans  = as.numeric(substr(A , 4, 4)) * (nchar(A) - 3)
ans  = as.numeric(substr(A , 5, 5)) * (nchar(A) - 4)
ans  = as.numeric(substr(A , 6, 6)) * (nchar(A) - 5)

This can be summarized as:

ans  = as.numeric(substr(A , i, i)) * (nchar(A) - i   1), i = 1, 2, ..., 6

Freshman

Loop as it is.

ans <- 0
for (i in 1:6) {
  ans <- ans   as.numeric(substr(A , i, i)) * (nchar(A) - i   1)
}
ans
#[1] 134

Sophomore

Pre-compute the result of substring and nchar, then index them in the loop.

N <- as.numeric(substring(A, 1:6, 1:6))
L <- nchar(A) - (1:6)   1
ans <- 0
for (i in 1:6) {
  ans <- ans   N[i] * L[i]
}
ans
#[1] 134

Junior

Replace the loop by sum.

N <- as.numeric(substring(A, 1:6, 1:6))
L <- nchar(A) - (1:6)   1
ans <- sum(N * L)
ans
#[1] 134

Senior

Replace sum by crossprod.

N <- as.numeric(substring(A, 1:6, 1:6))
L <- nchar(A) - (1:6)   1
ans <- c(crossprod(N, L))
ans
#[1] 134

CodePudding user response：

Your code can be considerably simpler and no loop is necessary. This may be what you are trying to accomplish:

A <- 883745
len <- nchar(A)
N <- as.numeric(unlist(strsplit(as.character(A), "")))
N
# [1] 8 8 3 7 4 5
N * (len - 0:5)
# [1] 48 40 12 21  8  5
sum(N * (len - 0:5))
# [1] 134