R使用ifelse函数为多个数据框创建新列

时间:2019-03-09 19:51:24

标签: r map-function

我想为多个数据框创建一个具有ifelse()条件的列。在这种情况下,数据帧是用于加密货币的3个时间序列数据。这是自动下载3个数据框的代码:

INotifyPropertyChanged

现在,我想在返回中检测到积极的过度反应(oR_pos)。我将过度反应定义为高于平均值+ 1标准偏差的值(返回值)。我也想针对1.5和2个标准差进行此操作。这是我想要的一种加密货币(比特币)的输出:

library(tidyverse)
library(crypto)

crypto_chart <- crypto_prices()%>% select(-id, -symbol,-price_btc, -`24h_volume_usd`,-available_supply, -total_supply,-max_supply, -percent_change_1h, -percent_change_24h, -percent_change_7d, -last_updated)%>% slice(1:3)

list_cryptocurrencies <-crypto_chart$name   

map(list_cryptocurrencies,
    function(x) crypto_history(x, start_date = '20150101', end_date = '20190303')%>%
      select(-slug, -symbol, -name, -`ranknow`))%>%
set_names(list_cryptocurrencies)%>%
list2env(envir = .GlobalEnv)

##Calculating return
map(mget(list_cryptocurrencies),
function(x) x %>% mutate(`return` =   (close-open)/open * 100))%>%
list2env(mget(list_cryptocurrencies), envir = .GlobalEnv)

现在我有3个新列,它们的反应过度(oR_pos)> 1sd; 1.5sd和2sd。

我已经尝试过此代码:

> Bitcoin
     date    open   close      return     oR_pos>1sd oR_pos>1.5sd oR_pos>2sd
1  2018-01-01 14112.2 13657.2  -3.2241607         NA           NA         NA
2  2018-01-02 13625.0 14982.1   9.9603670   9.960367     9.960367   9.960367
3  2018-01-03 14978.2 15201.0   1.4874952         NA           NA         NA
4  2018-01-04 15270.7 15599.2   2.1511784         NA           NA         NA
5  2018-01-05 15477.2 17429.5  12.6140387  12.614039    12.614039  12.614039
6  2018-01-06 17462.1 17527.0   0.3716621         NA           NA         NA
7  2018-01-07 17527.3 16477.6  -5.9889430         NA           NA         NA
8  2018-01-08 16476.2 15170.1  -7.9271919         NA           NA         NA
9  2018-01-09 15123.7 14595.4  -3.4931928         NA           NA         NA
10 2018-01-10 14588.5 14973.3   2.6376941         NA           NA         NA
11 2018-01-11 14968.2 13405.8 -10.4381288         NA           NA         NA
12 2018-01-12 13453.9 13980.6   3.9148500   3.914850           NA         NA

但是它不起作用。 有人可以帮我吗?

2 个答案:

答案 0 :(得分:1)

以下内容与您的预期功能紧密匹配,将所需的列添加到您的加密货币上,同时允许将所需的sd阈值作为参数传入以提高灵活性。顺便提一句,下面的解决方案根据OP使用>,但是您可能希望考虑从sd的移动+/-方向。可以改为使用以下解决方案:

col <- ifelse(returns > (r_mean+(r_sd*threshold)) | 
              returns < (r_mean-(r_sd*threshold)),
              returns,NA)

解决方案如下:

oR_pos_function <- function(returns,thresholds) {

  r_mean <- mean(returns,na.rm=T)
  r_sd <- sd(returns,na.rm=T)

  cols <- lapply(thresholds,function(threshold) {
    col <- ifelse(returns > (r_mean+(r_sd*threshold)),returns,NA)
    return(col)
  })
  cols <- as.data.frame(cols)
  names(cols) <- paste0("oR_pos>",thresholds,"sd")
  return(cols)  
}

new_cols <- oR_pos_function(returns=Bitcoin$return,thresholds=c(1,1.5,2))
Bitcoin <- cbind(Bitcoin,new_cols)

结果:

> head(Bitcoin[Bitcoin$date>="2018-01-01",])
           date    open    high     low   close      volume       market close_ratio spread     return oR_pos>1sd oR_pos>1.5sd oR_pos>2sd
1097 2018-01-01 14112.2 14112.2 13154.7 13657.2 10291200000 229119155396   0.5248042  957.5 -3.2241607         NA           NA         NA
1098 2018-01-02 13625.0 15444.6 13163.6 14982.1 16846600192 251377913955   0.7972381 2281.0  9.9603670   9.960367     9.960367   9.960367
1099 2018-01-03 14978.2 15572.8 14844.5 15201.0 16871900160 255080562912   0.4894961  728.3  1.4874952         NA           NA         NA
1100 2018-01-04 15270.7 15739.7 14522.2 15599.2 21783199744 261795321110   0.8845996 1217.5  2.1511784         NA           NA         NA
1101 2018-01-05 15477.2 17705.2 15202.8 17429.5 23840899072 292544135538   0.8898258 2502.4 12.6140387  12.614039    12.614039  12.614039
1102 2018-01-06 17462.1 17712.4 16764.6 17527.0 18314600448 294217423675   0.8043891  947.8  0.3716621         NA           NA         NA
> 

每个注释的替代项:

oR_pos_function <- function(coin_data,thresholds) {

  returns <- coin_data$return
  r_mean <- mean(returns,na.rm=T)
  r_sd <- sd(returns,na.rm=T)

  cols <- lapply(thresholds,function(threshold) {
    col <- ifelse(returns > (r_mean+(r_sd*threshold)),returns,NA)
    return(col)
  })
  cols <- as.data.frame(cols)
  names(cols) <- paste0("oR_pos>",thresholds,"sd")
  coin_data <- cbind(coin_data,cols)
  return(coin_data)  
}

答案 1 :(得分:0)

您可以使用#include <vector>添加任何此类字段

dplyr::mutate