快速编写多行代码的方法

时间：2016-01-17 10:28:09

标签： r

我正在研究R中的一个项目，与我之前的R项目相比，它至少相当重。代码在先前的列数据上使用多个ifelse语句，然后使用结果创建新列。由于我使用的数据是5分钟的时间范围，因此我必须为每5分钟的时间段编写一行新的代码。我的数据是从09:30到16:00，所以这是很多代码行，我的计算大约是75行。我的数据示例; 

    Date                  Open        High       Low         Close      doy
1   2015-09-21 09:30:00 164.6700    164.7100    164.3700    164.5300    264
2   2015-09-21 09:35:00 164.5300    164.9000    164.5300    164.6400    264
3   2015-09-21 09:40:00 164.6600    164.8900    164.6000    164.8900    264
4   2015-09-21 09:45:00 164.9100    165.0900    164.9100    164.9736    264
5   2015-09-21 09:50:00 164.9399    165.0980    164.8200    164.8200    264

然后将此数据过滤到这样的表格上; 

data <- structure(list(doy = c(264, 265, 266, 267, 268, 271, 272, 11,12, 13), Date = structure(c(1442824200, 1442910600, 1442997000,1443083400, 1443169800, 1443429000, 1443515400, 1452504600, 1452591000,1452677400), class = c("POSIXct", "POSIXt"), tzone = ""), Or_High = c(164.71,162.96, 163.38, 161.37, 163.91, 162.06, 160.22, 164.5, 165.23,165.84), OR_Low = c(164.37, 162.62, 162.98, 161.06, 163.57, 161.66,159.7, 164.06, 164.84, 165.4), HOD = c(165.56, 163.36, 163.38,162.24, 164.43, 162.06, 160.96, 164.5, 165.78, 165.84), LOD = c(165.22,163.1, 162.98, 161.95, 164.24, 161.66, 160.75, 164.06, 165.56,165.4), Close = c(164.92, 163.02, 162.58, 161.85, 162.94, 159.84,160.19, 163.83, 165.02, 161.38), Range = c(0.340000000000003,0.260000000000019, 0.400000000000006, 0.29000000000002, 0.189999999999998,0.400000000000006, 0.210000000000008, 0.439999999999998, 0.219999999999999,0.439999999999998), `A-val` = c(NA, NA, NA, NA, NA, NA, NA, 0.0673439999999994,0.0659639999999996, 0.0729499999999996), `A-up` = c(NA, NA, NA,NA, NA, NA, NA, 164.567344, 165.295964, 165.91295), `A-down` = c(NA,NA, NA, NA, NA, NA, NA, 163.992656, 164.774036, 165.32705), `09:35` = structure(c(NA,NA, NA, NA, NA, NA, NA, 0, 0, 0), .Dim = c(10L, 1L), .Dimnames = list(NULL, "Low")), `09:40` = structure(c(NA, NA, NA, NA, NA,NA, NA, -1, 1, 0), .Dim = c(10L, 1L), .Dimnames = list(NULL,"Low")), `09:45` = structure(c(NA, NA, NA, NA, NA, NA, NA,0, 1, 0), .Dim = c(10L, 1L), .Dimnames = list(NULL, "Low")),`09:50` = structure(c(NA, NA, NA, NA, NA, NA, NA, -1, 1,0), .Dim = c(10L, 1L), .Dimnames = list(NULL, "Low")), `09:55` = structure(c(NA,NA, NA, NA, NA, NA, NA, -1, 0, 0), .Dim = c(10L, 1L), .Dimnames = list(NULL, "Low")), `10:00` = structure(c(NA, NA, NA, NA,NA, NA, NA, -1, 0, 0), .Dim = c(10L, 1L), .Dimnames = list(NULL, "Low")), `10:05` = structure(c(NA, NA, NA, NA,NA, NA, NA, -1, 0, 0), .Dim = c(10L, 1L), .Dimnames = list(NULL, "Low")), `10:10` = structure(c(NA, NA, NA, NA,NA, NA, NA, -1, 0, 0), .Dim = c(10L, 1L), .Dimnames = list(NULL, "Low")), `10:15` = structure(c(NA, NA, NA, NA,NA, NA, NA, -2, 0, -1), .Dim = c(10L, 1L), .Dimnames = list(NULL, "Low")), `10:20` = structure(c(NA, NA, NA, NA,NA, NA, NA, 0, 0, -1), .Dim = c(10L, 1L), .Dimnames = list(NULL, "Low")), `10:25` = structure(c(NA, NA, NA, NA,NA, NA, NA, -2, -1, -1), .Dim = c(10L, 1L), .Dimnames = list(NULL, "Low")), `10:30` = structure(c(NA, NA, NA, NA,NA, NA, NA, 0, 0, -1), .Dim = c(10L, 1L), .Dimnames = list(NULL, "Low")), `10:35` = structure(c(NA, NA, NA, NA,NA, NA, NA, 0, 0, -1), .Dim = c(10L, 1L), .Dimnames = list(NULL, "Low")), `10:40` = structure(c(NA, NA, NA, NA,NA, NA, NA, 0, -1, -2), .Dim = c(10L, 1L), .Dimnames = list(NULL, "Low")), `10:45` = structure(c(NA, NA, NA, NA,NA, NA, NA, 0, -1, 0), .Dim = c(10L, 1L), .Dimnames = list(NULL, "Low")), `10:50` = structure(c(NA, NA, NA, NA,NA, NA, NA, -1, -1, -2), .Dim = c(10L, 1L), .Dimnames = list(NULL, "Low")), `10:55` = structure(c(NA, NA, NA, NA,NA, NA, NA, -1, -1, 0), .Dim = c(10L, 1L), .Dimnames = list(NULL, "Low"))), .Names = c("doy", "Date", "Or_High","OR_Low", "HOD", "LOD", "Close", "Range", "A-val", "A-up", "A-down","09:35", "09:40", "09:45", "09:50", "09:55", "10:00", "10:05","10:10", "10:15", "10:20", "10:25", "10:30", "10:35", "10:40","10:45", "10:50", "10:55"), row.names = c(1L, 2L, 3L, 4L, 5L,6L, 7L, 78L, 79L, 80L), class = "data.frame")

这就是代码行的样子; 

data[,14] <- ifelse(df %>% filter(hour(Date) == 09 & minute(Date) == 45) %>% select(Low) > data[,10], 1, ifelse(df %>% filter(hour(Date) == 09 & minute(Date) == 45) %>% select(High) < data[,11], -1, 0))

然后下一行代码看起来像; 

data[,15] <- ifelse(df %>% filter(hour(Date) == 09 & minute(Date) == 50) %>% select(Low) > data[,10], 1, ifelse(df %>% filter(hour(Date) == 09 & minute(Date) == 50) %>% select(High) < data[,11], -1, 0))

接下来就是这样等等; 

data[,16] <- ifelse(df %>% filter(hour(Date) == 09 & minute(Date) == 55) %>% select(Low) > data[,10], 1, ifelse(df %>% filter(hour(Date) == 09 & minute(Date) == 55) %>% select(High) < data[,11], -1, 0))

正如您在每个新代码行中看到的那样，只会更改代码的某些部分，例如用于求和的小时，分​​钟和列引用。也许下面的例子会更清楚。

实施例

colnames(data)[14] <- "09:45"
colnames(data)[15] <- "09:50" 
colnames(data)[16] <- "09:55"
colnames(data)[17] <- "10:00"
colnames(data)[18] <- "10:05"

在此代码中，无论如何都会改变[#col ref#]和时间而不用手动单独更改每行代码？我意识到复制和粘贴可以与记事本一起使用，但这仍然意味着编写单独的更改。我主要关注的不是写这篇文章的时间，而是人为输入错误的风险。

如果有人对如何做到这一点有任何提示或技巧，或者在不使用现有代码结构的多个if语句的情况下实现相同的其他方法，我将非常感谢您的帮助。这个问题与我发布的here上一个问题有关，可能会增加我想要实现的目标。

感谢。

1 个答案:

答案 0 :(得分：-1)

正如vanao veneri所说，最好使用文本编辑器快速编写批量代码。

我发现带有Text Pastry插件的Sublime 3使用insert nuns命令完全符合我的需要。

感谢您的帮助。

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