首页 > 解决方案 > 在 R 中平均两列部分匹配(大数据)

问题描述

我有一个矩阵,它有 562709 行和 803 列,对于列,它有 7 个元数据和 796 个复制数据,这些数据应该成对组合并取平均值。

我用谷歌搜索了同样的问题,但效果不佳,尤其是因为我的矩阵太大了,所以我想知道是否有人可以帮助我解决这个问题。

 [1] "seqnames"                                                          
 [2] "start"                                                             
 [3] "end"                                                               
 [4] "name"                                                              
 [5] "score"                                                             
 [6] "annotation"                                                        
 [7] "GC"                                                                
 [8] "ACCx_025FE5F8_885E_433D_9018_7AE322A92285_X034_S09_L133_B1_T1_PMRG"
 [9] "ACCx_025FE5F8_885E_433D_9018_7AE322A92285_X034_S09_L134_B1_T2_PMRG"
[10] "ACCx_2A5AE757_20D5_49B6_95FF_CAE08E8197A0_X012_S05_L033_B1_T1_P024"
[11] "ACCx_2A5AE757_20D5_49B6_95FF_CAE08E8197A0_X012_S05_L034_B1_T2_P025"
[12] "ACCx_3D0CD3BD_3960_46FB_92C3_777F11CCD0FC_X011_S06_L011_B1_T1_P024"
[13] "ACCx_3D0CD3BD_3960_46FB_92C3_777F11CCD0FC_X011_S06_L012_B1_T2_P026"
[14] "ACCx_4D0D43F5_D8F0_4735_92D5_F40E321C7A05_X010_S09_L065_B1_T1_P019"
[15] "ACCx_4D0D43F5_D8F0_4735_92D5_F40E321C7A05_X010_S09_L066_B1_T2_P020"
[16] "ACCx_81A262BD_3078_4BDB_8EB1_30DD6D7948C3_X027_S03_L081_B1_T1_P063"
[17] "ACCx_81A262BD_3078_4BDB_8EB1_30DD6D7948C3_X027_S03_L082_B1_T2_P067"
...
[800]"UCEC_C335297F_2D63_4973_9182_FA18C28E001E_X037_S04_L055_B1_T1_P088"
[801]"UCEC_C335297F_2D63_4973_9182_FA18C28E001E_X037_S04_L056_B1_T2_P089"
[802]"UCEC_D820B024_6B3B_4B5B_866E_F9A8139C270B_X039_S09_L113_B1_T1_P099"
[803]"UCEC_D820B024_6B3B_4B5B_866E_F9A8139C270B_X039_S09_L114_B1_T2_P098"

像上面一样,前 7 列不应该被修改,但是对于来自 8 的列,pair 应该合并为它们的平均值。(例如,第 8,9 列应该合并,而 10,11...)

标签: rbioinformatics

解决方案


获取交替列,将它们相加,然后除以 2:

# example data, 5 rows, 11 cols
x <- mtcars[1:5,  ]

cbind(
  # keep first 7 columns as is
  x[ 1:7 ],
  # then take alternating cols, add, and, divide by 2
  (x[ 8:11 ][, c(TRUE, FALSE) ] + x[ 8:11 ][, c(FALSE, TRUE) ]) / 2
  )

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