计算每个 ID 与前一天的差异

Question

Stock.Open <- rep(c(102.25,102.87,102.25,100.87,103.44,103.87,103.00),times=3)
Stock.Close <- rep(c(102.12,102.62,100.12,103.00,103.87,103.12,105.12), times=3)
Stock.id<-rep(1:3,each=7)                   
day<-rep(c(1:7),times=3)
df<-data.frame(day,Stock.Close,Stock.Open,Stock.id)

如何计算 stock.open 和 stock.close 每天与前一天的差异百分比。例如：我想计算第 1 天、第 2 天、第 2 天、第 3 天、第 3 天、第 4 天等之间 Stock.open 的百分比变化。

为每个 ID 执行相同的任务。

Answer 1

试试这个：

Stock.Open <- rep(c(102.25,102.87,102.25,100.87,103.44,103.87,103.00),times=3)
Stock.Close <- rep(c(102.12,102.62,100.12,103.00,103.87,103.12,105.12), times=3)
Stock.id<-rep(1:3,each=7)                   
day<-rep(c(1:7),times=3)
df<-data.frame(day,Stock.Close,Stock.Open,Stock.id)

#Subset stock data per day
temp <- subset(df, select = c(Stock.Open, Stock.Close, day, Stock.id))

#Change day and rename
temp$day <- temp$day + 1

require(plyr)
temp <- plyr::rename(temp, c("Stock.Open" = "Stock.Open.Pre", "Stock.Close" = "Stock.Close.Pre"))

#Merge back
df <- join(df, temp, by = c("day", "Stock.id"), type = "left")

#Compute difference
df$Stock.Open.Diff <- (df$Stock.Open / df$Stock.Open.Pre) - 1
df$Stock.Close.Diff <- (df$Stock.Close / df$Stock.Close.Pre) - 1

Answer 2

试试这个：

library(dplyr)

df %>%
  group_by(Stock.id) %>%
  arrange(day) %>%
  mutate(Change.Stock.Open = c(NA, diff(Stock.Open))/Stock.Open,
         Change.Stock.Close = c(NA, diff(Stock.Close))/Stock.Close)

# A tibble: 21 x 6
# Groups:   Stock.id [3]
     day Stock.Close Stock.Open Stock.id Change.Stock.Open Change.Stock.Close
   <int>       <dbl>      <dbl>    <int>             <dbl>              <dbl>
 1     1      102.12     102.25        1                NA                 NA
 2     1      102.12     102.25        2                NA                 NA
 3     1      102.12     102.25        3                NA                 NA
 4     2      102.62     102.87        1       0.006027024        0.004872345
 5     2      102.62     102.87        2       0.006027024        0.004872345
 6     2      102.62     102.87        3       0.006027024        0.004872345
 7     3      100.12     102.25        1      -0.006063570       -0.024970036
 8     3      100.12     102.25        2      -0.006063570       -0.024970036
 9     3      100.12     102.25        3      -0.006063570       -0.024970036
10     4      103.00     100.87        1      -0.013680976        0.027961165
# ... with 11 more rows

（与每只股票的第 1 天关联的值是 NA，因为没有前一天可比较）

Answer 3

另一种解决方案，这次只使用基础 R。

df$Stock.Close.Change <- ave(df$Stock.Close, df$Stock.id, FUN = function(x) c(NA, diff(x))/x)
df$Stock.Open.Change <- ave(df$Stock.Open, df$Stock.id, FUN = function(x) c(NA, diff(x))/x)
df <- df[order(df$day), ]
row.names(df) <- NULL
df
   day Stock.Close Stock.Open Stock.id Stock.Close.Change Stock.Open.Change
1    1      102.12     102.25        1                 NA                NA
2    1      102.12     102.25        2                 NA                NA
3    1      102.12     102.25        3                 NA                NA
4    2      102.62     102.87        1        0.004872345       0.006027024
5    2      102.62     102.87        2        0.004872345       0.006027024
6    2      102.62     102.87        3        0.004872345       0.006027024
7    3      100.12     102.25        1       -0.024970036      -0.006063570
8    3      100.12     102.25        2       -0.024970036      -0.006063570
9    3      100.12     102.25        3       -0.024970036      -0.006063570
10   4      103.00     100.87        1        0.027961165      -0.013680976
11   4      103.00     100.87        2        0.027961165      -0.013680976
12   4      103.00     100.87        3        0.027961165      -0.013680976
13   5      103.87     103.44        1        0.008375854       0.024845321
14   5      103.87     103.44        2        0.008375854       0.024845321
15   5      103.87     103.44        3        0.008375854       0.024845321
16   6      103.12     103.87        1       -0.007273080       0.004139790
17   6      103.12     103.87        2       -0.007273080       0.004139790
18   6      103.12     103.87        3       -0.007273080       0.004139790
19   7      105.12     103.00        1        0.019025875      -0.008446602
20   7      105.12     103.00        2        0.019025875      -0.008446602
21   7      105.12     103.00        3        0.019025875      -0.008446602