跳过空数据帧并产生输出答案

【问题标题】：skip the empty dataframes and produce the output跳过空数据帧并产生输出
【发布时间】：2015-09-04 08:07:37
【问题描述】：

 sessionInfo()
R version 3.2.0 (2015-04-16)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 7 x64 (build 7601) Service Pack 1

locale:
[1] LC_COLLATE=German_Germany.1252  LC_CTYPE=German_Germany.1252   
[3] LC_MONETARY=German_Germany.1252 LC_NUMERIC=C                   
[5] LC_TIME=German_Germany.1252    

attached base packages:
[1] grid      stats     graphics  grDevices utils     datasets  methods  
[8] base     

other attached packages:
[1] WriteXLS_3.5.1  tidyr_0.2.0     scales_0.2.4    gridExtra_0.9.1
[5] ggplot2_1.0.1   RPostgreSQL_0.4 DBI_0.3.1      

loaded via a namespace (and not attached):
 [1] Rcpp_0.11.6      assertthat_0.1   dplyr_0.4.1      digest_0.6.8    
 [5] MASS_7.3-40      plyr_1.8.2       gtable_0.1.2     magrittr_1.5    
 [9] stringi_0.4-1    lazyeval_0.1.10  reshape2_1.4.1   proto_0.3-10    
[13] tools_3.2.0      stringr_1.0.0    munsell_0.4.2    parallel_3.2.0  
[17] colorspace_1.2-6

library(RPostgreSQL)
    library(ggplot2)
    library(gridExtra)
    library(scales)
    library(tidyr)
    blue.bold.italic.16.text <- element_text(face = "bold", color = "black", size = 12)

假设有四台机器并行运行并生产产品。下面的每个数据帧(l1,l2,l3,l4) 表示每台机器每小时的片段数（实际上我使用 RPostgreSQL 从数据库收集数据，这些是它的外观示例）

l1 <- structure(list(hours = structure(c(1434081600, 1434085200, 1434088800, 
1434092400, 1434096000, 1434099600, 1434103200, 1434106800, 1434110400, 
1434114000, 1434117600, 1434121200, 1434124800, 1434128400, 1434132000, 
1434135600, 1434139200, 1434142800, 1434146400, 1434150000, 1434153600, 
1434157200, 1434160800, 1434164400), class = c("POSIXct", "POSIXt"
), tzone = ""), count = c(25, 29, 28, 32, 33, 13, 33, 29, 32, 
33, 27, 34, 25, 30, 13, 24, 26, 33, 40, 34, 26, 30, 22, 30)), .Names = c("hours", 
"count"), row.names = c(NA, 24L), class = "data.frame")

l2 <- structure(list(hours = structure(c(1434081600, 1434085200, 1434088800, 
1434092400, 1434096000, 1434099600, 1434103200, 1434106800, 1434110400, 
1434114000, 1434117600, 1434121200, 1434124800, 1434128400, 1434132000, 
1434135600, 1434139200, 1434142800, 1434146400, 1434150000, 1434153600, 
1434157200, 1434160800, 1434164400), class = c("POSIXct", "POSIXt"
), tzone = ""), count = c(25, 29, 28, 32, 33, 13, 33, 29, 32, 
33, 27, 34, 25, 30, 13, 24, 26, 33, 40, 34, 26, 30, 22, 30)), .Names = c("hours", 
"count"), row.names = c(NA, 24L), class = "data.frame")

l3 <- structure(list(hours = structure(c(1434081600, 1434085200, 1434088800, 
1434092400, 1434096000, 1434099600, 1434103200, 1434106800, 1434110400, 
1434114000, 1434117600, 1434121200, 1434124800, 1434128400, 1434132000, 
1434135600, 1434139200, 1434142800, 1434146400, 1434150000, 1434153600, 
1434157200, 1434160800, 1434164400), class = c("POSIXct", "POSIXt"
), tzone = ""), count = c(25, 29, 28, 32, 33, 13, 33, 29, 32, 
33, 27, 34, 25, 30, 13, 24, 26, 33, 40, 34, 26, 30, 22, 30)), .Names = c("hours", 
"count"), row.names = c(NA, 24L), class = "data.frame")

l4 <- structure(list(hours = structure(c(1434081600, 1434085200, 1434088800, 
1434092400, 1434096000, 1434099600, 1434103200, 1434106800, 1434110400, 
1434114000, 1434117600, 1434121200, 1434124800, 1434128400, 1434132000, 
1434135600, 1434139200, 1434142800, 1434146400, 1434150000, 1434153600, 
1434157200, 1434160800, 1434164400), class = c("POSIXct", "POSIXt"
), tzone = ""), count = c(25, 29, 28, 32, 33, 13, 33, 29, 32, 
33, 27, 34, 25, 30, 13, 24, 26, 33, 40, 34, 26, 30, 22, 30)), .Names = c("hours", 
"count"), row.names = c(NA, 24L), class = "data.frame")

# 这是我附加情节（输出）的脚本

df <- merge(l1,l2, by="hours")
df <- merge(df,l3, by="hours")
df <- merge(df,l4, by="hours")

colnames(df) <- c("hours","L 1","L 2","L 3","L 4")
pd <- gather(df, 'Ls', 'count', 2:5)

q <- ggplot(pd, aes(x = hours, y = count)) + geom_bar(stat = "identity") + theme(legend.position = "none")+
  xlab("Time") + ylab("No.Of Pecies") +
  ggtitle("my sample")+
  scale_y_continuous(breaks=seq(0,45, by = 5))+
  theme(axis.text = blue.bold.italic.16.text) +
  scale_x_datetime(breaks=date_breaks("2 hour"),minor_breaks=date_breaks("2 hour"),labels=date_format("%H")) + 
  theme(axis.text.x=element_text(angle=0))+
  facet_grid(~ Ls)

# 当所有 4 台机器都工作时 - 一切都很好，我将运行上面的脚本，我将得到所需的输出。

如果任何机器不工作并且我有一个空行的数据框..那么我在运行我的脚本文件时会出错。

 @ df <- merge(l1,l2, by="hours")
    df <- merge(df,l3, by="hours")
    df <- merge(df,l4, by="hours")
Error in fix.by(by.y, y) : 'by' must specify a uniquely valid column

和下一个错误在

pd <- gather(df, 'Ls', 'count', 2:5)

如何避免空数据帧并成功运行脚本以生成输出与任何机器正在运行（它是 2 或 3 或 4）

【问题讨论】：

带有“空行”的数据框是什么样的？包含零、NA、根本没有行？
顺便说一句，感谢您以有用的格式和良好的 MWE 提供数据

标签： r merge ggplot2 tidyr rpostgresql

【解决方案1】：

从报错信息来看，导致报错的data.frame既没有行也没有列，好像是NULL。因此，最简单的方法是检查这种情况，如果 data.frame 为 NULL，则创建一个可以是 merge()d 和 gather()ed 的虚拟对象。

我会做的（不是说这是最好的方法）是

# for easier looping, put your data.frames in a list
l <- list( l1, l2, l3, l4 )

# create a dummy that mimics the structure of your data.frames
dummy <- structure( list( hours = structure( c( Sys.time() ), 
                          class = c( "POSIXct", "POSIXt" ), tzone = ""),
                          count = c(0)), .Names = c("hours", "count"),
                          row.names = c(NA, 1L), class = "data.frame")

# check for empty data.frames and replace with dummy (will be NA)
for( i in 1:4 ) if( length( l[[ i ]] ) == 0 ) l[[ i ]] <- dummy

# merge
for( i in 2:4 ) l[[ 1 ]] <- merge( l[[ 1 ]], l[[ i ]], 
                                   by = "hours", all = TRUE )

# remove dummy and go back to your code
df <- l[[ 1 ]][ 1:24, ]
colnames( df ) <- c( "hours","L 1","L 2","L 3","L 4" )

还有改进的余地，但至少应该显示结果，无论机器是否正在运行：

l2 <- NULL

【讨论】：

@chanti - 如果你能接受答案会很好
我认为@vaettchen 的意思是接受堆栈溢出的答案。为此，请单击答案的复选标记（位于投票按钮下方）。
@chanti - 我的答案右上角有一个按钮，您可以将答案标记为“已接受”。这就是我的意思。当然，我更高兴听到你真的使用了我的想法……

【解决方案2】：

另一种选择是跳过所有合并并直接堆叠数据集。您只需先将 Ls 列添加到每个单独的数据集。

l1$Ls = "L 1"
l2$Ls = "L 2"
l3$Ls = "L 3"
l4$Ls = "L 4"

然后你可以使用，例如，来自 dplyr 的bind_rows 来制作你的长数据集pd。

bind_rows(l1, l2, l3, l4)

Source: local data frame [96 x 3]

                 hours count  Ls
1  2015-06-11 21:00:00    25 L 1
2  2015-06-11 22:00:00    29 L 1
3  2015-06-11 23:00:00    28 L 1
4  2015-06-12 00:00:00    32 L 1
5  2015-06-12 01:00:00    33 L 1
6  2015-06-12 02:00:00    13 L 1
7  2015-06-12 03:00:00    33 L 1
8  2015-06-12 04:00:00    29 L 1
9  2015-06-12 05:00:00    32 L 1
10 2015-06-12 06:00:00    33 L 1
..                 ...   ... ...

这种方法的好处是您绑定的对象之一可以是空的data.frame 或NULL，它仍然有效。

例如空data.frame:

l4.2 = data.frame()

bind_rows(l1, l2, l3, l4.2)

Source: local data frame [72 x 3]

                 hours count  Ls
1  2015-06-11 21:00:00    25 L 1
2  2015-06-11 22:00:00    29 L 1
3  2015-06-11 23:00:00    28 L 1
4  2015-06-12 00:00:00    32 L 1
5  2015-06-12 01:00:00    33 L 1
6  2015-06-12 02:00:00    13 L 1
7  2015-06-12 03:00:00    33 L 1
8  2015-06-12 04:00:00    29 L 1
9  2015-06-12 05:00:00    32 L 1
10 2015-06-12 06:00:00    33 L 1
..                 ...   ... ...

例如NULL:

l4.3 = NULL

bind_rows(l1, l2, l3, l4.3)

Source: local data frame [72 x 3]

                 hours count  Ls
1  2015-06-11 21:00:00    25 L 1
2  2015-06-11 22:00:00    29 L 1
3  2015-06-11 23:00:00    28 L 1
4  2015-06-12 00:00:00    32 L 1
5  2015-06-12 01:00:00    33 L 1
6  2015-06-12 02:00:00    13 L 1
7  2015-06-12 03:00:00    33 L 1
8  2015-06-12 04:00:00    29 L 1
9  2015-06-12 05:00:00    32 L 1
10 2015-06-12 06:00:00    33 L 1
..                 ...   ... ...

【讨论】：