我有一个文件列表,我正在尝试提取所有第1层的grd文件。在一个grep表达式中是否有这样的方法?
lof <- c("layer1_1.grd", "layer1_1.gri", "layer1_2.grd", "layer1_2.gri",
"layer1_3.grd", "layer1_3.gri", "layer1_4.grd", "layer1_4.gri",
"layer1_5.grd", "layer1_5.gri", "layer2_1.grd", "layer2_1.gri",
"layer2_2.grd", "layer2_2.gri", "layer2_3.grd", "layer2_3.gri",
"layer2_4.grd", "layer2_4.gri", "layer2_5.grd", "layer2_5.gri",
"layer3_1.grd", "layer3_1.gri", "layer3_2.grd", "layer3_2.gri",
"layer3_3.grd", "layer3_3.gri", "layer3_4.grd", "layer3_4.gri",
"layer3_5.grd", "layer3_5.gri", "layer4_1.grd", "layer4_1.gri",
"layer4_2.grd", "layer4_2.gri", "layer4_3.grd", "layer4_3.gri",
"layer4_4.grd", "layer4_4.gri", "layer4_5.grd", "layer4_5.gri")
我尝试了两个步骤:
list.of.files <- list.files(pattern = c("1_"))
list.of.files <- list.of.files[grep(".grd", list.of.files)]
有人能告诉我如何用grep一步完成这个任务吗?我天真地尝试将list()和c()传递给grep,但正如您所能想象的,它不起作用。
list.of.files <- list.files()
list.of.files <- list.of.files[grep(list("1_", ".grd"), list.of.files)]