2016-07-07 88 views
1

我從幾個重疊的NP /個體中找到了我試圖比較的基因分型數據。 正如你在下面的數據結構中看到的,e[1,2]e[2,3]有NA。現在我想用NA值代替d[1,2](1)和d[2,3](1)。根據另一個數據表中的NA替換一個數據表中的值

d <- structure(list(`100099681` = c(0L, 2L, 0L), `101666591` = c(1L, 1L, 0L), `102247652` = c(1L, 1L, 1L), `102284616` = c(0L, 1L, 0L), `103582612` = c(0L, 1L, 1L), `104344528` = c(2L, 1L, 0L),  `105729734` = c(1L, 0L, 1L), `109897137` = c(0L, 0L, 2L),  `112768301` = c(0L, 1L, 1L), `114724443` = c(1L, 1L, 1L),  `114826164` = c(1L, 0L, 1L), `115358770` = c(0L, 2L, 0L),  `115399788` = c(1L, 1L, 0L), `118669033` = c(0L, 1L, 1L),  `118875482` = c(2L, 1L, 0L), `119366362` = c(0L, 2L, 0L),  `119627971` = c(0L, 1L, 1L), `120295351` = c(0L, 2L, 0L),  `120998030` = c(0L, 0L, 2L)), .Names = c("100099681", "101666591", "102247652", "102284616", "103582612", "104344528", "105729734", "109897137", "112768301", "114724443", "114826164", "115358770", "115399788", "118669033", "118875482", "119366362", "119627971", "120295351", "120998030"), row.names = c("7:100038150_C", "7:100079759_T", "7:100256942_A"), class = "data.frame") 
> d 
#    100099681 101666591 102247652 102284616 103582612 104344528 105729734 109897137 112768301 114724443 114826164 115358770 115399788 118669033 118875482 119366362 119627971 120295351 120998030 
#7:100038150_C   0   1   1   0   0   2   1   0   0   1   1   0   1   0   2   0   0   0  0 
#7:100079759_T   2   1   1   1   1   1   0   0   1   1   0   2   1   1   1   2   1   2  0 
#7:100256942_A   0   0   1   0   1   0   1   2   1   1   1   0   0   1   0   0   1   0  2 

e<- structure(list(`100099681` = c(1L, 1L, 0L), `101666591` = c(NA, 1L, 1L), `102247652` = c(0L, NA, 0L), `102284616` = c(1L, 1L, 0L), `103582612` = c(1L, 0L, 1L), `104344528` = c(1L, 0L, 1L),  `105729734` = c(0L, 0L, 1L), `109897137` = c(1L, 1L, 0L),  `112768301` = c(0L, 1L, 1L), `114724443` = c(0L, 2L, 0L),  `114826164` = c(0L, 0L, 2L), `115358770` = c(0L, 0L, 2L),  `115399788` = c(0L, 2L, 0L), `118669033` = c(0L, 0L, 2L),  `118875482` = c(0L, 1L, 1L), `119366362` = c(2L, 1L, 0L),  `119627971` = c(0L, 1L, 1L), `120295351` = c(0L, 2L, 0L),  `120998030` = c(0L, 2L, 1L)), .Names = c("100099681", "101666591", "102247652", "102284616", "103582612", "104344528", "105729734", "109897137", "112768301", "114724443", "114826164", "115358770", "115399788", "118669033", "118875482", "119366362", "119627971", "120295351", "120998030"), row.names = c("7:100038150_C", "7:100079759_T", "7:100256942_A"), class = "data.frame") 
> e 
#    100099681 101666591 102247652 102284616 103582612 104344528 105729734 109897137 112768301 114724443 114826164 115358770 115399788 118669033 118875482 119366362 119627971 120295351 120998030 
#7:100038150_C   1  NA   0   1   1   1   0   1   0   0   0   0   0   0   0   2   0   0   0 
#7:100079759_T   1   1  NA   1   0   0   0   1   1   2   0   0   2   0   1   1   1   2   2 
#7:100256942_A   0   1   0   0   1   1   1   0   1   0   2   2   0   2   1   0   1   0   1 

因此我預期的產出將是

> expected_d 
#    100099681 101666591 102247652 102284616 103582612 104344528 105729734 109897137 112768301 114724443 114826164 115358770 115399788 118669033 118875482 119366362 119627971 120295351 120998030 
#7:100038150_C   0   NA   1   0   0   2   1   0   0   1   1   0   1   0   2   0   0   0  0 
#7:100079759_T   2   1   NA   1   1   1   0   0   1   1   0   2   1   1   1   2   1   2  0 
#7:100256942_A   0   0   1   0   1   0   1   2   1   1   1   0   0   1   0   0   1   0  2 

到目前爲止,我已經得到了這一點;

g <- which(is.na(e), arr.ind=TRUE) 
> g 
#    row col 
#7:100038150_C 1 2 
#7:100079759_T 2 3 

然後嘗試使用apply功能由「TEST」更換位置(或NA爲此事)

apply(g, 1, function(x){ 
    e[x[1], x[2]] <- "TEST" } 
) 
#> apply(g, 1, function(x){ e[x[1], x[2]] <- "TEST" }) 
#7:100038150_C 7:100079759_T 
#  "TEST"  "TEST"  

我將在幾百萬行/列運行這段代碼所以速度將是一個問題。 預先感謝您:)

回答

2

根據你的方法的另一種方式,

d[which(is.na(e), arr.ind = T)] <- NA 
4

我們可以嘗試做

NA^(is.na(e))*d 

如果記憶是一個問題

d[] <- Map(function(x,y) NA^(is.na(y))* x, d, e) 
+3

哇。這很聰明。 –

+0

與Steven一致,在0/1/2的數據集上像一個魅力一樣工作,但不適用於包含字符的列:'在NA ^(is.na(adf))中的錯誤* tdf:非數字參數二元運算符 執行停止' – Bas

+1

@Bas您的示例沒有顯示任何非數字列。 – akrun

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