I just cant figure it out how to create a vector in which the strings are constant but the numbers are not. For example:
我只是想不通如何创建一个矢量，其中字符串是常量但数字不是。例如：
c("raster[1]","raster[2]","raster[3]")
I'd like to use something like seq(raster[1],raster[99], by=1), but this does not work.
我想使用像seq（raster [1]，raster [99]，by = 1）这样的东西，但这不起作用。
Thanks in advance.
提前致谢。
2 个解决方案
#1
17
The sprintf function should also work:
sprintf函数也应该工作：
rasters <- sprintf("raster[%s]",seq(1:99))
head(rasters)
[1] "raster[1]" "raster[2]" "raster[3]" "raster[4]" "raster[5]" "raster[6]"
As suggested by Richard Scriven, %d is more efficient than %s. So, if you were working with a longer sequence, it would be more appropriate to use:
正如Richard Scriven所建议的，％d比％s更有效。因此，如果您使用更长的序列，则使用更合适：
rasters <- sprintf("raster[%d]",seq(1:99))
#2
10
We can do
我们可以做的
paste0("raster[", seq(1:6), "]")
# [1] "raster[1]" "raster[2]" "raster[3]" "raster[4]" "raster[5]" "raster[6]"
#1
17
The sprintf function should also work:
sprintf函数也应该工作：
rasters <- sprintf("raster[%s]",seq(1:99))
head(rasters)
[1] "raster[1]" "raster[2]" "raster[3]" "raster[4]" "raster[5]" "raster[6]"
As suggested by Richard Scriven, %d is more efficient than %s. So, if you were working with a longer sequence, it would be more appropriate to use:
正如Richard Scriven所建议的，％d比％s更有效。因此，如果您使用更长的序列，则使用更合适：
rasters <- sprintf("raster[%d]",seq(1:99))
#2
10
We can do
我们可以做的
paste0("raster[", seq(1:6), "]")
# [1] "raster[1]" "raster[2]" "raster[3]" "raster[4]" "raster[5]" "raster[6]"