Question

我有一个具有2行30406列的大型数据框。我需要计算给定列（匹配）的两行中出现0的次数以及给定列中另一行而不是另一行中出现0的次数（不匹配）。

我认为，如果我仅遍历所有内容并比较每列，那么考虑到> 30k列，这将花费很长时间

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UINavigationBar.appearance(whenContainedInInstancesOf: [QLPreviewController.self]).titleTextAttributes = [NSAttributedString.Key.foregroundColor: UIColor.white]
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present(quickLookController, animated: true, completion: nil)

输出

head(to_compare)[1:5]
     bin:82154:182154 bin:82154:282154 bin:82154:382154 
bin:82154:482154
1-D1.txt                0                1                2                
0
1-D2.txt                1                1                1                
1
     bin:82154:582154
1-D1.txt                0
1-D2.txt                0

Answer 1

您可以将colSums用于矢量化解决方案：

set.seed(123)
df <- as.data.frame(matrix(round(runif(50, 0, 2)), nrow = 2))

# Match 
sum(colSums(df==0) == 2)
[1] 2

# No match
sum(colSums(df==0) == 1)
[1] 8

Answer 2

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Answer 3

另一种非常简单的方法是先将列切换为行，然后使用rowSums：

#Create sample df
df <- data.frame(col1 = c(0,1), col2 = c(1,0), col3 = c(1,1), col4 = c(0,2), col5 = c(3,0), col6 = c(0,0))

#Convert columns to rows
df_long <- t(df)

#Count number of 0s in every row and show in table of 0, 1 or 2 zeros
table(rowSums(df_long == 0))

0 1 2 
1 4 1

Answer 4

set.seed(7)
n <- 30406
to_compare <- data.frame(matrix(floor(runif(n*2, 0, 3)), nrow = 2))

table(colSums(to_compare==0))
#    0     1     2 
#13519 13513  3374 
#
#0..no zero in column (13519)
#1..one row in column has a zero (13513)
#2..both rows in column are zero (3374)

system.time(table(colSums(to_compare==0)))
#       User      System verstrichen 
#      0.332       0.000       0.330

快速比较两行的方法

4 个答案: