如何仅在矩阵或热图中显示具有某些列名的值

Question

我有一个矩阵，它在行和列之间匹配了样本。我已经从这个矩阵中生成了一个很好的热图，但是为了清楚起见，我只想在列名根据 grep 与行名匹配时显示那些样本 数据：

    structure(c(447, 439, 365, 359, 342, 341, 382, 356, 364, 309, 
295, 527, 410, 415, 323, 291, 292, 266, 323, 337, 309, 366, 331, 
284, 414, 425, 316, 419, 420, 350, 301, 319, 293, 335, 360, 312, 
341, 313, 284, 328, 327, 312, 344, 361, 324, 323, 328, 357, 426, 
412, 338, 309, 291, 493, 406, 427, 328, 307, 283, 555, 317, 302, 
501, 312, 289, 480, 423, 419, 336, 299, 274, 241, 286, 262, 221, 
418, 415, 352, 310, 291, 572, 445, 428, 361, 320, 317, 543, 324, 
291, 479, 320, 301, 538, 327, 308, 553, 423, 433, 332, 347, 345, 
324, 347, 345, 324, 410, 413, 351, 340, 322, 558, 331, 321, 581, 
306, 285, 525, 407, 397, 327, 325, 310, 565, 309, 314, 461, 339, 
335, 304, 369, 342, 349, 292, 301, 477, 321, 302, 450, 325, 314, 
582, 275, 252, 456, 296, 291, 526, 401, 414, 326, 335, 316, 582
), .Dim = c(3L, 51L), .Dimnames = list(c("AH_026T", "AH_058T", 
"AH_084T"), c("AH_026C", "AH_026C.1", "AH_026C.2", "AH_084C", 
"AH_086C", "AH_086C.1", "AH_086C.2", "AH_086C.3", "AH_088C", 
"AH_094C", "AH_094C.1", "AH_094C.2", "AH_094C.3", "AH_094C.4", 
"AH_094C.5", "AH_094C.6", "AH_094C.7", "AH_096C", "AH_100C", 
"AH_100C.1", "AH_127C", "AH_133C", "ED_008C", "ED_008C.1", "ED_008C.2", 
"ED_008C.3", "ED_016C", "ED_031C", "ED_036C", "GS_001C", "QE_062C", 
"RS_010C", "RS_027C", "RS_027C.1", "RS_027C.2", "SH_051C", "ST_014C", 
"ST_014C.1", "ST_020C", "ST_024C", "ST_033C", "ST_034C", "ST_034C.1", 
"ST_034C.2", "ST_035C", "ST_036C", "ST_040C", "WG_002C", "WG_005C", 
"WG_006C", "WG_019C")))

grep 应该只匹配前 6 个字符。如果列名和行名不匹配，则可以是 NA 或零。

我试过了：

tryout<- subset(res_matrixMatchRes,colnames(res_matrixMatchRes)==rownames(res_matrixMatchRes))

但它不起作用。我想我可能需要使用 lapply 来遍历列并查看它是否与每一行匹配，但后来我被难住了

Answer 1

一种使用sapply的方式：

#use sapply to check the names
#this checks the 6 first characters of the row.names against the
#first 6 characters of the column names. If they match it produces TRUE
#else it produces FALSE
new_mat <- t(sapply(row.names(df),
                    function(x) substr(x, 1, 6) == substr(colnames(df), 1, 6)))

#then you just subset the original df using the logical matrix created previously 
df[!new_mat] <- 0

输出：

> df
        AH_026C AH_026C.1 AH_026C.2 AH_084C AH_086C AH_086C.1 AH_086C.2 AH_086C.3 AH_088C AH_094C AH_094C.1 AH_094C.2 AH_094C.3
AH_026T     447       359       382       0       0         0         0         0       0       0         0         0         0
AH_058T       0         0         0       0       0         0         0         0       0       0         0         0         0
AH_084T       0         0         0     527       0         0         0         0       0       0         0         0         0
#result truncated...