使用三个其他给定变量，迭代或按行滚动计算两个同时发生的变量

本文介绍了使用三个其他给定变量，迭代或按行滚动计算两个同时发生的变量的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

名为 cras 的数据集看起来像 -

The dataset named crass looks like -

> dput(crass)
structure(list(WT_TRADE_PRICE = c(3801, 3801, 3801, 3797, 3797,
3796.2125, 3800, 3797, 3795.09523809524, 3794, 3793, 3793, 3793.8,
3794.72, 3793.02777777778, 3789, 3790, 3788, 3788, 3788), min = c(3801,
3801, 3801, 3797, 3797, 3795, 3800, 3797, 3794, 3794, 3793, 3793,
3793, 3794, 3790, 3789, 3790, 3788, 3788, 3788), max = c(3801,
3801, 3801, 3797, 3797, 3800, 3800, 3797, 3797, 3794, 3793, 3793,
3794, 3797, 3794, 3789, 3790, 3788, 3788, 3788), Bid = c(3801,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA), Ask = c(3802, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), row.names = c(NA, -20L
), class = c("tbl_df", "tbl", "data.frame"))

# A tibble: 20 x 5
   WT_TRADE_PRICE   min   max   Bid   Ask
            <dbl> <dbl> <dbl> <dbl> <dbl>
 1          3801   3801  3801  3801  3802
 2          3801   3801  3801    NA    NA
 3          3801   3801  3801    NA    NA
 4          3797   3797  3797    NA    NA
 5          3797   3797  3797    NA    NA
 6          3796.  3795  3800    NA    NA
 7          3800   3800  3800    NA    NA
 8          3797   3797  3797    NA    NA
 9          3795.  3794  3797    NA    NA
10          3794   3794  3794    NA    NA
11          3793   3793  3793    NA    NA
12          3793   3793  3793    NA    NA
13          3794.  3793  3794    NA    NA
14          3795.  3794  3797    NA    NA
15          3793.  3790  3794    NA    NA
16          3789   3789  3789    NA    NA
17          3790   3790  3790    NA    NA
18          3788   3788  3788    NA    NA
19          3788   3788  3788    NA    NA
20          3788   3788  3788    NA    NA

可以看出，两个变量ask &bid 只有初始值，需要使用以下逻辑迭代填充.

As can be seen, two variables ask & bid have only initial values which need to be filled iteratively using the following logic.

如果 WT_TRDAE_PRICE 是 >= 比之前的 Ask OR WT_TRADE_PRICE 值> 将检查 > 是否比之前的 bid 和 ask 的平均值 - 然后将当前的 ask 设置为相等到当前行 max 变量，bid 将设置为等于之前的 bid 值.
否则，当前的 ask 将被设置为之前的 ask 值 &bid 到 max.

If WT_TRDAE_PRICE is >= than previous value of Ask OR WT_TRADE_PRICE will be checked whether > than mean of previous bid and ask - then current ask will be set equal to current row max variable and bid will be set equal to previous bid value.
Else , current ask will be set to previous ask value & bid to max.

伪代码-

if(WT_TRADE_PRICE >= L(Ask) | WT_TRADE_PRICE > (L(Bid)+L(Ask))/2)
{
  Bid = L(Bid), Ask = max
}
else
{
  Bid = min, Ask = L(Ask)
}

最终输出-

SNo.	WT_TRADE_PRICE	分钟	最大	出价	询问
1	3801	3801	3801	3801	3802
2	3801	3801	3801	3801	3802
3	3801	3801	3801	3801	3802
4	3797	3797	3797	3797	3802
5	3797	3797	3797	3797	3802
6	3796.	3795	3800	3795	3802
7	3800	3800	3800	3795	3800
8	3797	3797	3797	3797	3800
9	3795.	3794	3797	3794	3800
10	3794	3794	3794	3794	3800
11	3793	3793	3793	3793	3800
12	3793	3793	3793	3793	3800
13	3794.	3793	3794	3793	3800
14	3795.	3794	3797	3793	3797
15	3793.	3790	3794	3790	3797
16	3789	3789	3789	3789	3797
17	3790	3790	3790	3790	3797
18	3788	3788	3788	3788	3797
19	3788	3788	3788	3788	3797
20	3788	3788	3788	3788	3797

推荐答案

这种情况下，我们需要同时生成两个输出列；并在三个输入的帮助下迭代.所以 purrr::accumulate 通常基于一个输入对一个输出起作用，而 purrr::accumulate2() 对 2 个输入起作用，再次产生一个输出.因此，我的 accumulate 策略如下:-

In this case, we require to generate two output columns simultaneously; and iteratively with the help of three inputs. So purrr::accumulate normally works on one output based on one input, whereas purrr::accumulate2() works on 2 inputs for again one output. So instead, my strategy for accumulate is as under:-

将三个输入列重新排列成行式小标题，这样三列输入中的每一列现在都是一列.为此，我生成了一个虚拟列 id，以便每一行都转换为一个 tibble.
为此我使用了 tidyr::nest_by()
再次为输出，我通过累加生成了一个小标题而不是一个向量.
最后，我使用 tidyr::unnnest_wider()

Re-arrange three input columns into row-wise tibbles each, so that each of three columns inputs are now a single column. For this, I generated a dummy column id so that each row is converted to a tibble each.
I used tidyr::nest_by() for this
again for output I generated a tibble instead of a vector through accumulate.
Lastly I converted both tibbles back to their original shapes by using tidyr::unnnest_wider()

crass[1:3] %>%
  nest_by(id = row_number()) %>%
  ungroup() %>%
  mutate(new = accumulate(data,
                          .init = list(Bid = 3801, Ask = 3802),
                          ~ tibble(Bid = ifelse(.y$WT_TRADE_PRICE >= min(.x$Ask, (.x$Ask + .x$Bid)/2),
                                                     .x$Bid,
                                                     .y$min),
                                        Ask = ifelse(.y$WT_TRADE_PRICE >= min(.x$Ask, (.x$Ask + .x$Bid)/2),
                                                     .y$max,
                                                     .x$Ask))
                          )[-1]) %>%
  unnest_wider(data) %>%
  unnest_wider(new)

# A tibble: 20 x 6
      id WT_TRADE_PRICE   min   max   Bid   Ask
   <int>          <dbl> <dbl> <dbl> <dbl> <dbl>
 1     1          3801   3801  3801  3801  3802
 2     2          3801   3801  3801  3801  3802
 3     3          3801   3801  3801  3801  3802
 4     4          3797   3797  3797  3797  3802
 5     5          3797   3797  3797  3797  3802
 6     6          3796.  3795  3800  3795  3802
 7     7          3800   3800  3800  3795  3800
 8     8          3797   3797  3797  3797  3800
 9     9          3795.  3794  3797  3794  3800
10    10          3794   3794  3794  3794  3800
11    11          3793   3793  3793  3793  3800
12    12          3793   3793  3793  3793  3800
13    13          3794.  3793  3794  3793  3800
14    14          3795.  3794  3797  3794  3800
15    15          3793.  3790  3794  3790  3800
16    16          3789   3789  3789  3789  3800
17    17          3790   3790  3790  3790  3800
18    18          3788   3788  3788  3788  3800
19    19          3788   3788  3788  3788  3800
20    20          3788   3788  3788  3788  3800

较早修订的 for 循环


for(i in 2:nrow(crass)){
    if(crass[i, 1] >= min(crass[i-1, 5], (crass[i-1, 4] + crass[i-1, 5])/2)){
    crass[i, 5] <- crass[i, 3]
    crass[i, 4] <- crass[i-1, 4]
  } else {
    crass[i, 4] <- crass[i, 2]
    crass[i, 5] <- crass[i-1, 5]
  }
}

crass
# A tibble: 20 x 5
   WT_TRADE_PRICE   min   max   Bid   Ask
            <dbl> <dbl> <dbl> <dbl> <dbl>
 1          3801   3801  3801  3801  3802
 2          3801   3801  3801  3801  3802
 3          3801   3801  3801  3801  3802
 4          3797   3797  3797  3797  3802
 5          3797   3797  3797  3797  3802
 6          3796.  3795  3800  3795  3802
 7          3800   3800  3800  3795  3800
 8          3797   3797  3797  3797  3800
 9          3795.  3794  3797  3794  3800
10          3794   3794  3794  3794  3800
11          3793   3793  3793  3793  3800
12          3793   3793  3793  3793  3800
13          3794.  3793  3794  3793  3800
14          3795.  3794  3797  3794  3800
15          3793.  3790  3794  3790  3800
16          3789   3789  3789  3789  3800
17          3790   3790  3790  3790  3800
18          3788   3788  3788  3788  3800
19          3788   3788  3788  3788  3800
20          3788   3788  3788  3788  3800

crass for 循环运行前

# A tibble: 20 x 5
   WT_TRADE_PRICE   min   max   Bid   Ask
            <dbl> <dbl> <dbl> <dbl> <dbl>
 1          3801   3801  3801  3801  3802
 2          3801   3801  3801    NA    NA
 3          3801   3801  3801    NA    NA
 4          3797   3797  3797    NA    NA
 5          3797   3797  3797    NA    NA
 6          3796.  3795  3800    NA    NA
 7          3800   3800  3800    NA    NA
 8          3797   3797  3797    NA    NA
 9          3795.  3794  3797    NA    NA
10          3794   3794  3794    NA    NA
11          3793   3793  3793    NA    NA
12          3793   3793  3793    NA    NA
13          3794.  3793  3794    NA    NA
14          3795.  3794  3797    NA    NA
15          3793.  3790  3794    NA    NA
16          3789   3789  3789    NA    NA
17          3790   3790  3790    NA    NA
18          3788   3788  3788    NA    NA
19          3788   3788  3788    NA    NA
20          3788   3788  3788    NA    NA

这篇关于使用三个其他给定变量，迭代或按行滚动计算两个同时发生的变量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！