For this you can use the modulo operator, i.e. %%
. Take for example:
> 322%%8
[1] 2
which tells you that after dividing 322 by 8, 2 remains, i.e. 320 is exactly 40 times 8, leaving 2.
In you example we can use %%
combined with subsetting to get the the multiples of 8. Remember that %%
yields 0 for exact multiples of 8:
input = 1:1000
multiple_of_8 = (input %% 8) == 0
head(multiple_of_8)
[1] FALSE FALSE FALSE FALSE FALSE FALSE
length(multiple_of_8)
[1] 1000
also note that %%
is a vectorized operation, i.e. of the left hand side is a vector, the result will also be a vector. The multiple_of_8
vector now contains 1000 logicals stating if that particular element of input
is an exact multiple of 8. Using that logical vector to subset get's you the result you need:
input[multiple_of_8]
[1] 8 16 24 32 40 48 56 64 72 80 88 96 104 112 120
[16] 128 136 144 152 160 168 176 184 192 200 208 216 224 232 240
[31] 248 256 264 272 280 288 296 304 312 320 328 336 344 352 360
[46] 368 376 384 392 400 408 416 424 432 440 448 456 464 472 480
[61] 488 496 504 512 520 528 536 544 552 560 568 576 584 592 600
[76] 608 616 624 632 640 648 656 664 672 680 688 696 704 712 720
[91] 728 736 744 752 760 768 776 784 792 800 808 816 824 832 840
[106] 848 856 864 872 880 888 896 904 912 920 928 936 944 952 960
[121] 968 976 984 992 1000
or more compactly:
input[(input %% 8) == 0]
与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…