Question

我正在寻找一个例程，它将向量舍入“必要”的数字位数，以便所有元素仍然可以区分。我的第一次尝试看起来像这样：

discr.round <- function(x) {
  digits <- ceiling(-min(log10(diff(sort(x)))))
  round(x, digits)
}

discr.round(c(12.336, 12.344))
# [1] 12.336 12.344
discr.round(c(12.336, 12.347))
# [1] 12.34 12.35
discr.round(c(10, 25, 39))
# [1] 10 20 40
discr.round(c(1.2345679, 1.2345681))
# [1] 1.234568 1.234568
discr.round(c(1.23456789, 1.23456791))
# [1] 1.234568 1.234568

我需要使用“少量”小数位来格式化浮点数的向量。

如何调用这种舍入？在基础R或包中是否存在此标准或类似逻辑的“标准”实现？

编辑：我需要配对，但代码应该推广到任意长度的矢量。如果数字完全相同，则不需要区分它们。

Answer 1

options(digits=12) # Part of the problem is that your default digits = 7,
                   #   so you won't see any digits beyond that.
                   #  If you put too many (e.g, 22), you'll get floating point imprecision

#  Note the behavior for rounding off a 5 under ?round:
#  "IEC 60559 standard is expected to be used, ‘go to the even digit’."

# digits needed to distinguish between two nonequal elements
diff.dec.places <- function(x, y){
  delta <- abs(x - y)
  dec.place <- -ceiling(log10(abs(delta))) # your idea here was correct
  print(paste0("The elements (",x," & ",y,") differ in the ",
               10^-dec.place,"'s place."))
  if(round(x, dec.place)==round(y, dec.place)){
    print("But we add another sig. figure so they do not round to the same number.")
    dec.place <- dec.place + 1
  } 
  print(paste("The elements will round to",round(x, dec.place),'&',round(y, dec.place)))
  dec.place
}


# formula for arbitrary number of elements:
discr.round <- function(x){

  #- Find minimum-magnitude difference and which elements possess it: 
  #-   Create upper triangle of difference matrix of elements of vector with itself
  d <- abs(outer(x,x,"-"))
  d[lower.tri(d, diag=T)] <- NA
  #-   Return the smallest-magnitude difference and indices of elements forming it
  m <- min(d, na.rm=T)     
  if(m != 0){
    #- Round to number of dec places required to distinguish the closest elements
    e <- x[which(d==m, arr.ind=T)]
    round(x, diff.dec.places(e[1],e[2]))
  }
  else{
    print("Closest elements are equal.") 
    x
  }
}


discr.round(c(12.336, 12.344))
# [1] 12.336 12.344
discr.round(c(12.336, 12.347))
# [1] 12.34 12.35
discr.round(c(10, 25, 39))
# [1] 10 20 40
discr.round(c(1.2345679, 1.2345681))
# [1] 1.2345679 1.2345681
discr.round(c(1.23456789, 1.23456791))
# [1] 1.23456789 1.23456791

舍入向量以使所有结果元素不同

1 个答案: