我有一个data.table,它有许多行对,它们的q.val值不同。对于这些对中的每一对,我希望选择具有较小值的行。换句话说,我想从DT1转到DT2 (见下文)。在data.table包或其他包中有没有简单的方法可以做到这一点?
DT1
cluster q.val
c1 8.68E-03
c1 1.00E+00
c2 4.53E-05
c2 1.00E+00
c3 2.46E-03
c3 1.00E+00
c4 4.18E-05
c4 1.00E+00
c5 1.00E+00
c5 3.98E-05
c6 1.00E+00
c6 4.71E-06
DT2
cluster q.val
c1 8.68E-03
c2 4.53E-05
c3 2.46E-03
c4 4.18E-05
c5 3.98E-05
c6 4.71E-06根据我通过原始问题编辑的答案,发布我真实表格的一部分
cluster pathway q.val
c1 Adrenergic signaling in cardiomyocytes -3.01E-06
c1 Adrenergic signaling in cardiomyocytes -1.80E+00
c2 Adrenergic signaling in cardiomyocytes -5.07E-06
c2 Adrenergic signaling in cardiomyocytes -1.30E+00
c3 Adrenergic signaling in cardiomyocytes -1.46E-06
c3 Adrenergic signaling in cardiomyocytes -2.32E+00
c4 Adrenergic signaling in cardiomyocytes -1.60E-05
c4 Adrenergic signaling in cardiomyocytes -1.75E+00
c5 Adrenergic signaling in cardiomyocytes 2.58E+00
c5 Adrenergic signaling in cardiomyocytes 2.53E-06
c6 Adrenergic signaling in cardiomyocytes 3.54E+00
c6 Adrenergic signaling in cardiomyocytes 8.74E-08
c7 Adrenergic signaling in cardiomyocytes -4.85E-02
c7 Adrenergic signaling in cardiomyocytes -3.98E-03
c8 Adrenergic signaling in cardiomyocytes 9.73E-01
c8 Adrenergic signaling in cardiomyocytes 3.44E-05
c1 Aldosterone synthesis and secretion -3.01E-06
c1 Aldosterone synthesis and secretion -1.64E+00
c2 Aldosterone synthesis and secretion -5.07E-06
c2 Aldosterone synthesis and secretion -1.49E+00
c3 Aldosterone synthesis and secretion -1.46E-06
c3 Aldosterone synthesis and secretion -1.85E+00
c4 Aldosterone synthesis and secretion -1.60E-05
c4 Aldosterone synthesis and secretion -1.40E+00
c5 Aldosterone synthesis and secretion 2.58E+00
c5 Aldosterone synthesis and secretion 2.53E-06
c6 Aldosterone synthesis and secretion 3.45E+00
c6 Aldosterone synthesis and secretion 8.74E-08
c7 Aldosterone synthesis and secretion -1.28E-02
c7 Aldosterone synthesis and secretion -1.42E-02
c8 Aldosterone synthesis and secretion 4.24E-01
c8 Aldosterone synthesis and secretion 3.44E-05发布于 2018-03-03 04:32:39
DT1[, min(q.val), by = c("cluster", "pathway")]data.table中的基本语法允许您使用"by“语句来区分函数(在本例中为”min“)的应用位置。值得注意的是,只有当每个集群没有多个不同的路径时,这才会起作用。如果有,则每个集群将有多个行。
https://stackoverflow.com/questions/49076898
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