Question

我有data.frame我要删除连续几天的行。例如，我有以下data.frame（头），其名称为sell_tv，我想删除具有连续日期的行。在这种特殊情况下，我想删除第5行，如第5行和第5行。 6有连续的日期。

    Date Open High  Low Close  Sell.TV   Buy.TV
1 2015-04-08 2207 2204 2165  2166 4.038113 3.083603
2 2015-03-16 2214 2215 2172  2198 4.041986 3.087017
3 2015-03-05 2343 2364 2320  2324 4.023689 3.081034
4 2015-01-27 2171 2182 2151  2178 4.021998 3.070200
5 2015-01-23 2234 2244 2222  2230 4.032086 3.061206
6 2015-01-22 2278 2282 2242  2246 4.037248 3.095450

我为此编写了以下代码，但获得了：

****"Error in if (sell_tv$Date[i] == sell_tv$Date[i + 1] + 1) { :   missing value where TRUE/FALSE needed"****

代码：

for( i in 1:nrow(sell_tv))
{
  if (sell_tv$Date[i] == sell_tv$Date[i+1] + 1 )
  {
    new_sell<- sell_tv[-i,]  
  }       
  else
  {
    new_sell<- sell_tv[,]
  }
  i= i+1      
}

感谢任何帮助！

Answer 1

正如我在评论中所说，您可以保留循环并保存应在变量中删除的行数，或者您可以尝试一次获取行号：

to_delete <- which(sell_tv$Date[-nrow(sell_tv)]==sell_tv$Date[-1]+1) #5
new_sell <- sell_tv[-to_delete, ]

new_sell
        # Date Open High  Low Close  Sell.TV   Buy.TV
# 1 2015-04-08 2207 2204 2165  2166 4.038113 3.083603
# 2 2015-03-16 2214 2215 2172  2198 4.041986 3.087017
# 3 2015-03-05 2343 2364 2320  2324 4.023689 3.081034
# 4 2015-01-27 2171 2182 2151  2178 4.021998 3.070200
# 6 2015-01-22 2278 2282 2242  2246 4.037248 3.095450

数据

sell_tv <- structure(list(Date = structure(c(16533, 16510, 16499, 16462, 16458, 16457), class = "Date"), Open = c(2207L, 2214L, 2343L, 2171L, 2234L, 2278L), High = c(2204L, 2215L, 2364L, 2182L, 2244L, 2282L), Low = c(2165L, 2172L, 2320L, 2151L, 2222L, 2242L), Close = c(2166L, 2198L, 2324L, 2178L, 2230L, 2246L), Sell.TV = c(4.038113, 4.041986, 4.023689, 4.021998, 4.032086, 4.037248), Buy.TV = c(3.083603, 3.087017, 3.081034, 3.0702, 3.061206, 3.09545)), .Names = c("Date", "Open", "High", "Low", "Close", "Sell.TV", "Buy.TV"), row.names = c("1", "2", "3", "4", "5", "6"), class = "data.frame")

Answer 2

此解决方案可用于sell_tv数据框的日期列中的唯一日期和重复日期

sell_tv = read.table("myfile.txt", sep = "\t", header = TRUE, stringsAsFactors = FALSE)

print(sell_tv)
#         Date Open High  Low Close  Sell.TV   Buy.TV
# 1 2015-04-08 2207 2204 2165  2166 4.038113 3.083603
# 2 2015-03-16 2214 2215 2172  2198 4.041986 3.087017
# 3 2015-03-05 2343 2364 2320  2324 4.023689 3.081034
# 4 2015-01-27 2171 2182 2151  2178 4.021998 3.070200
# 5 2015-01-23 2234 2244 2222  2230 4.032086 3.061206
# 6 2015-01-22 2278 2282 2242  2246 4.037248 3.095450

#add duplicate date
sell_tv[3,1] = "2015-01-23"

print(sell_tv)
# Date Open High  Low Close  Sell.TV   Buy.TV
# 1 2015-04-08 2207 2204 2165  2166 4.038113 3.083603
# 2 2015-03-16 2214 2215 2172  2198 4.041986 3.087017
# 3 2015-01-23 2343 2364 2320  2324 4.023689 3.081034
# 4 2015-01-27 2171 2182 2151  2178 4.021998 3.070200
# 5 2015-01-23 2234 2244 2222  2230 4.032086 3.061206
# 6 2015-01-22 2278 2282 2242  2246 4.037248 3.095450

date_str = sell_tv$Date

to_delete = c()

for(i in date_str){
  a1 = which(unlist(lapply(date_str, function(x) as.numeric(difftime(x, i))))== 1)
  if(length(a1) > 0){
    to_delete = c(to_delete, a1) 
    } else
      next
}

sell_tv = sell_tv[-to_delete,]

输出：

print(sell_tv)
        Date Open High  Low Close  Sell.TV   Buy.TV
1 2015-04-08 2207 2204 2165  2166 4.038113 3.083603
2 2015-03-16 2214 2215 2172  2198 4.041986 3.087017
4 2015-01-27 2171 2182 2151  2178 4.021998 3.070200
6 2015-01-22 2278 2282 2242  2246 4.037248 3.095450

Answer 3

在日期使用带有diff运算符的逻辑索引：

sell_tv[ c(9999,diff(sell_tv$Date)) != -1, ]

我们只是将一些sentinel值添加到diff(...)

的输出中

如果你想排除＆＃39;之前或之后的那一天，那么布尔值 - 不是％运算符中的％：

sell_tv[ ! (c(9999,diff(sell_tv$Date)) %in% c(-1,+1)), ]

删除包含连续日期的行

3 个答案: