r - 如何根据元素情况删除一行？

Question

我有一个像下面这样的数据框

V1        V2       V3        V4      V5
ATP1A2  ATP1A1  220948_s_at LARS2   204016_at
CAPRIN1 ATP1A2              LARS2   
ATP1A1  ATP1A2  203296_s_at MTHFD1  202309_at
CBX3    ATP1B1  201243_s_at PRKCSH  200707_at
AUP1    ATP1B1              PRKCSH  214080_x_at
LARS2   AUP1    220525_s_at PRKCSH  
MTHFD1  AUP1                VDAC2   211662_s_at
VDAC2   B3GNT3  204856_at   VDAC2   
PRKCSH  B3GNT3          
ATP1B1  CAPRIN1 200722_s_at     
B3GNT3  CAPRIN1 200723_s_at     
        CAPRIN1         
        CBX3    200037_s_at

我想从 V2 中删除 V3 为空的那些元素，或者从 V4 中删除 V5 为空的元素，例如在 V3 中，第二个元素为空，因此应该删除 V3 和 V2 的第二个元素。输出结果如下所示。

     V1        V2       V3         V4      V5
    ATP1A2  ATP1A1   220948_s_at   LARS2   204016_at
    CAPRIN1 ATP1A2   203296_s_at   MTHFD1  202309_at                  
    ATP1A1  ATP1B1   201243_s_at   PRKCSH  200707_at
    CBX3    AUP1     220525_s_at   PRKCSH  214080_x_at
    AUP1    B3GNT3   204856_at     VDAC2   211662_s_at
    LARS2   CAPRIN1  200722_s_at   
    MTHFD1  CAPRIN1  200723_s_at   
    VDAC2   CBX3     200037_s_at  
    PRKCSH          
    ATP1B1     
    B3GNT3       

我可以检测到如何在每一列中找到那些空字符串，但问题是如何根据它删除两列行并对所有数据框执行此操作

# to find the empty elements 
Rcolumn3 <-which(df$V3=="")
Rcolumn5 <-which(df$V5=="") 

df<- structure(list(V1 = structure(c(3L, 7L, 2L, 8L, 5L, 9L, 10L, 
12L, 11L, 4L, 6L, 1L, 1L), .Label = c("", "ATP1A1", "ATP1A2", 
"ATP1B1", "AUP1", "B3GNT3", "CAPRIN1", "CBX3", "LARS2", "MTHFD1", 
"PRKCSH", "VDAC2"), class = "factor"), V2 = structure(c(1L, 2L, 
2L, 3L, 3L, 4L, 4L, 5L, 5L, 6L, 6L, 6L, 7L), .Label = c("ATP1A1", 
"ATP1A2", "ATP1B1", "AUP1", "B3GNT3", "CAPRIN1", "CBX3"), class = "factor"), 
    V3 = structure(c(9L, 1L, 6L, 5L, 1L, 8L, 1L, 7L, 1L, 3L, 
    4L, 1L, 2L), .Label = c("", "200037_s_at", "200722_s_at", 
    "200723_s_at", "201243_s_at", "203296_s_at", "204856_at", 
    "220525_s_at", "220948_s_at"), class = "factor"), V4 = structure(c(2L, 
    2L, 3L, 4L, 4L, 4L, 5L, 5L, 1L, 1L, 1L, 1L, 1L), .Label = c("", 
    "LARS2", "MTHFD1", "PRKCSH", "VDAC2"), class = "factor"), 
    V5 = structure(c(4L, 1L, 3L, 2L, 6L, 1L, 5L, 1L, 1L, 1L, 
    1L, 1L, 1L), .Label = c("", "200707_at", "202309_at", "204016_at", 
    "211662_s_at", "214080_x_at"), class = "factor")), .Names = c("V1", 
"V2", "V3", "V4", "V5"), class = "data.frame", row.names = c(NA, 
-13L))

获取数据

Answer 1

这是一种方式:

#convert to character
df[] <- lapply(df, as.character)

#convert the equivalent V2 or V4 columns' rows into '' 
a <- which(df$V3=='')
b <- which(df$V5=='')

df$V2[a] <- ''
df$V4[b] <- ''

此时我已经编写了一个将空白推到向量/列的末尾/底部的函数:

push_up <- function(vec){
   c(vec[-which(vec=='')], vec[which(vec=='')])
}

并将其应用到 df:

df[] <- lapply(df, push_up)

输出:

> df
        V1      V2          V3     V4          V5
1   ATP1A2  ATP1A1 220948_s_at  LARS2   204016_at
2  CAPRIN1  ATP1A2 203296_s_at MTHFD1   202309_at
3   ATP1A1  ATP1B1 201243_s_at PRKCSH   200707_at
4     CBX3    AUP1 220525_s_at PRKCSH 214080_x_at
5     AUP1  B3GNT3   204856_at  VDAC2 211662_s_at
6    LARS2 CAPRIN1 200722_s_at                   
7   MTHFD1 CAPRIN1 200723_s_at                   
8    VDAC2    CBX3 200037_s_at                   
9   PRKCSH                                       
10  ATP1B1                                       
11  B3GNT3                                       
12                                               
13                

或 df[df$V1 != '',] 删除第 12,13 行