Question

我想绘制一个热图并仅对行进行聚类（即该tydf1中的基因）。另外，是否要使热图的列标签顺序与df（即tydf1）中的顺序相同？

样本数据

df1 <- structure(list(Gene = c("AA", "PQ", "XY", "UBQ"), X_T0_R1 = c(1.46559502, 0.220140568, 0.304127515, 1.098842127), X_T0_R2 = c(1.087642983, 0.237500819, 0.319844338, 1.256624804), X_T0_R3 = c(1.424945196, 0.21066267, 0.256496284, 1.467120048), X_T1_R1 = c(1.289943948, 0.207778662, 0.277942721, 1.238400358), X_T1_R2 = c(1.376535013, 0.488774258, 0.362562315, 0.671502431), X_T1_R3 = c(1.833390311, 0.182798731, 0.332856558, 1.448757569), X_T2_R1 = c(1.450753714, 0.247576125, 0.274415259, 1.035410946), X_T2_R2 = c(1.3094609, 0.390028842, 0.352460646, 0.946426593), X_T2_R3 = c(0.5953716, 1.007079177, 1.912258811, 0.827119776), X_T3_R1 = c(0.7906009, 0.730242116, 1.235644748, 0.832287694), X_T3_R2 = c(1.215333041, 1.012914813, 1.086362205, 1.00918082), X_T3_R3 = c(1.069312467, 0.780421013, 1.002313082, 1.031761442), Y_T0_R1 = c(0.053317766, 3.316414959, 3.617213894, 0.788193798), Y_T0_R2 = c(0.506623748, 3.599442788, 1.734075583, 1.179462912), Y_T0_R3 = c(0.713670106, 2.516735845, 1.236204882, 1.075393433), Y_T1_R1 = c(0.740998252, 1.444496448, 1.077023349, 0.869258744), Y_T1_R2 = c(0.648231834, 0.097957459, 0.791438659, 0.428805547), Y_T1_R3 = c(0.780499252, 0.187840968, 0.820430227, 0.51636582), Y_T2_R1 = c(0.35344654, 1.190274584, 0.401845911, 1.223534348), Y_T2_R2 = c(0.220223951, 1.367784148, 0.362815405, 1.102117612), Y_T2_R3 = c(0.432856978, 1.403057729, 0.10802472, 1.304233845), Y_T3_R1 = c(0.234963735, 1.232129062, 0.072433381, 1.203096462), Y_T3_R2 = c(0.353770497, 0.885122768, 0.011662112, 1.188149743), Y_T3_R3 = c(0.396091395, 1.333921747, 0.192594116, 1.838029829), Z_T0_R1 = c(0.398000559, 1.286528398, 0.129147097, 1.452769794), Z_T0_R2 = c(0.384759325, 1.122251177, 0.119475721, 1.385513609), Z_T0_R3 = c(1.582230097, 0.697419716, 2.406671502, 0.477415567), Z_T1_R1 = c(1.136843842, 0.804552001, 2.13213228, 0.989075996), Z_T1_R2 = c(1.275683837, 1.227821594, 0.31900326, 0.835941568), Z_T1_R3 = c(0.963349308, 0.968589683, 1.706670339, 0.807060135), Z_T2_R1 = c(3.765036263, 0.477443352, 1.712841882, 0.469173869), Z_T2_R2 = c(1.901023385, 0.832736132, 2.223429427, 0.593558769), Z_T2_R3 = c(1.407713024, 0.911920317, 2.011259223, 0.692553388), Z_T3_R1 = c(0.988333629, 1.095130142, 1.648598854, 0.629915612), Z_T3_R2 = c(0.618606729, 0.497458337, 0.549147265, 1.249492088), Z_T3_R3 = c(0.429823986, 0.471389536, 0.977124788, 1.136635484)), row.names = c(NA, -4L ), class = c("data.table", "data.frame"))

使用的脚本

library(dplyr) 
library(stringr) 
library(tidyr) 
gdf1 <- gather(df1, "group", "Expression", -Gene) 
gdf1$tgroup <- apply(str_split_fixed(gdf1$group, "_", 3)[, c(1, 2)], 
                     1, paste, collapse ="_")

library(dplyr) 
tydf1 <- gdf1 %>% 
  group_by(Gene, tgroup) %>% 
  summarize(expression_mean = mean(Expression)) %>% 
  spread(., tgroup, expression_mean)

正在使用＃1热图脚本

library(tidyverse)
tydf1 <- tydf1 %>% 
  as.data.frame() %>% 
  column_to_rownames(var=colnames(tydf1)[1]) 

library(gplots) 
library(vegan) 
randup.m <- as.matrix(tydf1) 
scaleRYG <- colorRampPalette(c("red","yellow","darkgreen"), 
                             space = "rgb")(30) 
data.dist <- vegdist(randup.m, method = "euclidean") 
row.clus <- hclust(data.dist, "aver") 
heatmap.2(randup.m, Rowv = as.dendrogram(row.clus), 
          dendrogram = "row", col = scaleRYG, margins = c(7,10), 
          density.info = "none", trace = "none", lhei = c(2,6), 
          colsep = 1:3, sepcolor = "black", sepwidth = c(0.001,0.0001), 
          xlab = "Identifier", ylab = "Rows")

正在使用＃2热图脚本

df2 <- as.matrix(tydf1[, -1]) 
heatmap(df2)

我还要添加一个颜色键。

Answer 1

我仍然不清楚所需的输出是多少。有一些注意事项：

您不需要使用vegdist()来计算hclust()通话的距离矩阵。因为如果您检查all(vegdist(randup.m, method = "euclidian") == dist(randup.m))，它将返回TRUE;
在您的Colv = F调用中指定heatmap.2()将阻止对列进行重新排序（默认为TRUE）；
也许最好按行缩放数据（请参阅未注释的行）；
您对heatmap.2()的调用会返回带有颜色键的热图。

总结一下-在您的第一个脚本中，您只是错过了Colv = F参数，而在稍作调整后，它看起来像这样：

heatmap.2(randup.m, 
          Rowv = as.dendrogram(row.clus),
          Colv = F,
          dendrogram = "row", 
          #scale = "row", 
          col = scaleRYG,
          density.info = "none",
          trace = "none",
          srtCol = -45,
          adjCol = c(.1, .5),
          xlab = "Identifier",
          ylab = "Rows"
          )

但是我仍然不确定-是您需要的吗？

HeatMap：如何仅对行进行聚类，并使热图的列标签的顺序与df中的相同？

1 个答案: