r - 添加向量的连续元素直到一个值

Question

我想计算一个向量中连续元素的最小数量，当添加(连续)时将小于给定值。

例如在下面的向量中

ev<-c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 2.7, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3.27, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 370.33, 1375.4, 
1394.03, 1423.8, 1360, 1269.77, 1378.8, 1350.37, 1425.97, 1423.6, 
1363.4, 1369.87, 1365.5, 1294.97, 1362.27, 1117.67, 1026.97, 
1077.4, 1356.83, 565.23, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 356.83, 
973.5, 0, 240.43, 1232.07, 1440, 1329.67, 1096.87, 1331.37, 1305.03, 
1328.03, 1246.03, 1182.3, 1054.53, 723.03, 1171.53, 1263.17, 
1200.37, 1054.8, 971.4, 936.4, 968.57, 897.93, 1099.87, 876.43, 
1095.47, 1132, 774.4, 1075.13, 982.57, 947.33, 1096.97, 929.83, 
1246.9, 1398.2, 1063.83, 1223.73, 1174.37, 1248.5, 1171.63, 1280.57, 
1183.33, 1016.23, 1082.1, 795.37, 900.83, 1159.2, 992.5, 967.3, 
1440, 804.13, 418.17, 559.57, 563.87, 562.97, 1113.1, 954.87, 
883.8, 1207.1, 1046.83, 995.77, 803.93, 1036.63, 946.9, 887.33, 
727.97, 733.93, 979.2, 1176.8, 1241.3, 1435.6)

连续添加时(如向量中的顺序)总计为 20000 的最少元素数是多少

为了更清楚，我需要以下内容:从 ev[1] 开始并连续添加到 20000。记录为达到 20000 而必须添加的元素数作为 r[1]。然后从 ev[2] 开始并添加到 20000 等等。将您必须添加到 20000 的元素数记录为 r[2]。对 ev 的整个长度执行此操作。然后返回 min(r)

例如

j<-c(1, 2, 3, 5, 7, 9, 2) .

我想要连续添加时元素的最小数量可以让我们说 >20。这应该是 3 (5+7+9)

非常感谢

Answer 1

好吧，我试一试:这个会找到最小数字序列的长度加起来等于或高于 max。它没有声称速度很快，但它具有 O(2n) 时间复杂度:-)

我让它同时返回起始索引和长度。

f <- function(x, max=10) {
  s <- 0
  len <- Inf
  start <- 1
  j <- 1
  for (i in seq_along(x)) {
     s <- s + x[i]
     while (s >= max) {
       if (i-j+1 < len) {
         len <- i-j+1
         start <- j
       }
       s <- s - x[j]
       j <- j + 1
     }
  }

  list(start=start, length=len)
  # uncomment the line below if you don't need the start index...
  #len
}

r <- f(ev, 20000) # list(start=245, length=15)
sum(ev[seq(r$start, len=r$length)]) # 20275.42

# Test speed:
x <- sin(1:1e6)

system.time( r <- f(x, 1.9) ) # 1.54 secs

# Compile the function makes it 9x faster...
g <- compiler::cmpfun(f)
system.time( r <- g(x, 1.9) ) # 0.17 secs