Question

我最近发布了一个似乎不太清楚https://stackoverflow.com/questions/28857329/how-to-make-a-threshold-for-a-given-data-frame的问题，因为有一个答案我无法删除它。我在这里提出了一个明确的问题我有一个如下所示的数据框

M<- structure(list(V1 = structure(c(6L, 2L, 4L, 8L, 7L, 3L, 1L, 5L
), .Label = c("203797_at", "205217_at", "211488_s_at", "211900_x_at", 
"213959_s_at", "217077_s_at", "219884_at", "220473_s_at"), class = "factor"), 
    V2 = structure(c(8L, 6L, 4L, 2L, 7L, 1L, 5L, 3L), .Label = c("202498_s_at", 
    "203313_s_at", "204407_at", "207022_s_at", "212030_at", "218566_s_at", 
    "220926_s_at", "222204_s_at"), class = "factor"), V3 = structure(c(7L, 
    2L, 8L, 1L, 3L, 6L, 4L, 5L), .Label = c("201368_at", "201502_s_at", 
    "202211_at", "202422_s_at", "206542_s_at", "212902_at", "215509_s_at", 
    "215716_s_at"), class = "factor"), V4 = structure(c(2L, 4L, 
    7L, 6L, 5L, 1L, 3L, 8L), .Label = c("203736_s_at", "204442_x_at", 
    "205882_x_at", "207317_s_at", "208138_at", "213731_s_at", 
    "215743_at", "218513_at"), class = "factor"), V5 = structure(c(7L, 
    5L, 1L, 4L, 2L, 3L, 8L, 6L), .Label = c("202052_s_at", "203809_s_at", 
    "206319_s_at", "206590_x_at", "208382_s_at", "216133_at", 
    "219736_at", "221818_at"), class = "factor")), .Names = c("V1", 
"V2", "V3", "V4", "V5"), class = "data.frame", row.names = c(NA, 
-8L))

我有一个名为T

的专栏

T<- structure(list(V1 = structure(c(5L, 8L, 6L, 1L, 2L, 7L, 4L, 3L
), .Label = c("203797_at", "205217_at", "211488_s_at", "211900_x_at", 
"213959_s_at", "217077_s_at", "219884_at", "220473_s_at"), class = "factor")), .Names = "V1", row.names = c(8L, 
4L, 1L, 7L, 2L, 5L, 3L, 6L), class = "data.frame")

我想检查T的第一个字符是否在M的第一列中，如果它是1，如果它不是则为零。

# empty matrix with the same size az M 
output <- matrix (0,nrow(M),ncol(M))

输出应该如下所示

    [,1] [,2] [,3] [,4] [,5]
[1,]    1    0    0    0    0
[2,]    1    0    0    0    0
[3,]    1    0    0    0    0
[4,]    1    0    0    0    0
[5,]    1    0    0    0    0
[6,]    1    0    0    0    0
[7,]    1    0    0    0    0
[8,]    1    0    0    0    0

如果一个角色存在1，如果不是零

Answer 1

不漂亮，但有效......

sapply(1:ncol(M),function(i) sapply(T,function(t) t %in% M[,i]))*1

删除'* 1'以获得逻辑矩阵。

Answer 2

这是一个解决方案：

> apply(M, 2, function(col)as.numeric(col%in%t(T)))

     V1 V2 V3 V4 V5
[1,]  1  0  0  0  0
[2,]  1  0  0  0  0
[3,]  1  0  0  0  0
[4,]  1  0  0  0  0
[5,]  1  0  0  0  0
[6,]  1  0  0  0  0
[7,]  1  0  0  0  0
[8,]  1  0  0  0  0

如何在列和数据框之间找到匹配？

2 个答案: