我的列表中包含具有此格式值的列：mx6q:/sys/class/gpio/gpio200# echo "out" > direction <-- set the direction as write(out)
mx6q:/sys/class/gpio/gpio200# cat direction < -- just to verify
out
mx6q:/sys/class/gpio/gpio200# echo 1 > value < -- set value as 1(high)
mx6q:/sys/class/gpio/gpio200# cat value
1
mx6q:/sys/class/gpio/gpio200# echo 0 > value < -- set value Low again
mx6q:/sys/class/gpio/gpio200# cat value
0
mx6q:/sys/class/gpio/gpio200# cat direction
out
mx6q:/sys/class/gpio/gpio200# echo "in" > direction < -- change the direction to read the pin
mx6q:/sys/class/gpio/gpio200# cat direction
in
mx6q:/sys/class/gpio/gpio200# cat value < -- what i am expecting here is 0 (last set value)
1
。此列的类型称为11/1/2017 0:00。如何将此日期更改为integer？
由于
答案 0 :(得分：0)
x <- lubridate::mdy_hm(x)
x <- sprintf("%s/%s", lubridate::month(x), lubridate::day(x))