当你必须保持跑步平衡时，有没有比for循环更好的解决方案？

我有一个包含数百万行的大型数据帧。它是时间序列数据。例如：

dates <- c(1,2,3)
purchase_price <- c(5,2,1)
income <- c(2,2,2)
df <- data.frame(dates=dates,price=purchase_price,income=income)

有没有更快的解决方案

---

谢谢

正如Paul指出的，一些迭代是必要的。一个实例和上一个点之间存在依赖关系

但是，只有在进行购买时才会发生依赖关系（请阅读：您只需要在..时重新计算余额）。因此，您可以“批量”进行迭代

尝试下面的方法，通过确定下一行中哪一行有足够的余额进行购买。然后，它在单个调用中处理前面的所有行，然后从该点开始

library(data.table)
DT <- as.data.table(df)

## Initial Balance
b.init <- 2

setattr(DT, "Starting Balance", b.init)

## Raw balance for the day, regardless of purchase
DT[, balance := b.init + cumsum(income)]
DT[, buying  := FALSE]

## Set N, to not have to call nrow(DT) several times
N   <- nrow(DT)

## Initialize
ind <- seq(1:N)

# Identify where the next purchase is
while(length(buys <- DT[ind, ind[which(price <= balance)]]) && min(ind) < N) {
  next.buy <- buys[[1L]] # only grab the first one
  if (next.buy > ind[[1L]]) {
    not.buys <- ind[1L]:(next.buy-1L)
    DT[not.buys, buying := FALSE]
  }
  DT[next.buy, `:=`(buying  = TRUE
                  , balance = (balance - price)
                  ) ]

  # If there are still subsequent rows after 'next.buy', recalculate the balance
  ind <- (next.buy+1) : N
#  if (N > ind[[1]]) {  ## So that
    DT[ind, balance := cumsum(income) + DT[["balance"]][[ ind[[1]]-1L]] ]
#  }
}
# Final row needs to be outside of while-loop, or else will buy that same item multiple times
if (DT[N, !buying && (balance > price)])
  DT[N, `:=`(buying  = TRUE, balance = (balance - price)) ]

库（data.table）
DT对于容易用循环表示的问题，我越来越相信Rcpp是正确的解决方案。这是相对的，你可以很自然地表达loop-y算法

以下是使用Rcpp解决问题的方法：

#include <Rcpp.h>
using namespace Rcpp;

// [[Rcpp::export]]
List purchaseWhenPossible(NumericVector date, NumericVector income, 
                          NumericVector price, double init_balance = 0) {  
  int n = date.length();
  NumericVector balance(n);
  LogicalVector buy(n);

  for (int i = 0; i < n; ++i) {
    balance[i] = ((i == 0) ? init_balance : balance[i - 1]) + income;

    // Buy it if you can afford it
    if (balance[i] >= price[i]) {
      buy[i] = true;
      balance[i] -= price[i];
    } else {
      buy[i] = false;
    }

  }

  return List::create(_["buy"] = buy, _["balance"] = balance);
}

/*** R

# Copying input data from Ricardo
df <- data.frame(
  dates = 1:6,
  income = rep(2, 6),
  price = c(5, 2, 3, 5, 2, 1)
)

out <- purchaseWhenPossible(df$dates, df$income, df$price, 3)
df$balance <- out$balance
df$buy <- out$buy

*/

#包括
使用名称空间Rcpp；
//[[Rcpp:：导出]]
列出可能的采购（数字矢量日期、数字矢量收入、，
数值向量价格，双初始值=0{
int n=date.length（）；
数值矢量平衡（n）；
LogicalVector购买（n）；
对于（int i=0；i=价格[i]）{
买[我]=真的；
余额[i]=价格[i]；
}否则{
买[我]=假；
}
}
退货清单：：创建（[“购买”]=购买，[“余额”]=余额）；
}
/***R
#从Ricardo复制输入数据
df我删除了我的答案，这只是下一步尝试的提示。我同意有太多的相互依赖，不能简单地使用
cumsum（）
。也许有人会顺便来看看谁有解决方案。如果
价格高于当前的
余额，为什么要将
余额设置为零？试试xts package函数apply.daily.@Fernando这仍然会在这里的每一行上循环。应该是
购买金额=行$price*（余额>=行$price）
## Show output
{
  print(DT)
  cat("Starting Balance was", attr(DT, "Starting Balance"), "\n")
}


## Starting with 3: 
   dates price income balance buying
1:     1     5      2       0   TRUE
2:     2     2      2       0   TRUE
3:     3     3      2       2  FALSE
4:     4     5      2       4  FALSE
5:     5     2      2       4   TRUE
6:     6     1      2       5   TRUE
Starting Balance was 3

## Starting with 2: 
   dates price income balance buying
1:     1     5      2       4  FALSE
2:     2     2      2       4   TRUE
3:     3     3      2       3   TRUE
4:     4     5      2       0   TRUE
5:     5     2      2       0   TRUE
6:     6     1      2       1   TRUE
Starting Balance was 2


# I modified your original data slightly, for testing
df <- rbind(df, df)
df$dates <- seq_along(df$dates)
df[["price"]][[3]] <- 3

#include <Rcpp.h>
using namespace Rcpp;

// [[Rcpp::export]]
List purchaseWhenPossible(NumericVector date, NumericVector income, 
                          NumericVector price, double init_balance = 0) {  
  int n = date.length();
  NumericVector balance(n);
  LogicalVector buy(n);

  for (int i = 0; i < n; ++i) {
    balance[i] = ((i == 0) ? init_balance : balance[i - 1]) + income;

    // Buy it if you can afford it
    if (balance[i] >= price[i]) {
      buy[i] = true;
      balance[i] -= price[i];
    } else {
      buy[i] = false;
    }

  }

  return List::create(_["buy"] = buy, _["balance"] = balance);
}

/*** R

# Copying input data from Ricardo
df <- data.frame(
  dates = 1:6,
  income = rep(2, 6),
  price = c(5, 2, 3, 5, 2, 1)
)

out <- purchaseWhenPossible(df$dates, df$income, df$price, 3)
df$balance <- out$balance
df$buy <- out$buy

*/