Suppose that X has 20 elements. After the command: REBIN X XOUT 2 XOUT[1] = X[ 1] + X[ 2] XOUT[2] = X[ 3] + X[ 4] XOUT[3] = X[ 5] + X[ 6] ... XOUT[10] = X[19] + X[20]
After the command: REBIN [1:1000] X 3 X[1] = 1+2+3 X[2] = 4+5+6 X[3] = 7+8+9 ... X[333] = 997+998+999 A warning will be given that length of [1:1000] is not evenly divisible by 3 and so the last bin is incomplete.