An example of calculating "FizzBuzz" in R using dplyr and magrittr

fizzbuzz.md

Apparently, there is a simple problem called Fizz buzz which is sometimes used to identify competent programmers. A good opportunity to practice some dplyr and magrittr tricks.

library(dplyr)
library(magrittr)
library(knitr)

1:100 %>%
  data.frame %>%
  set_names("num") %>%
  tbl_df %>%
  mutate(fizz=num %>% mod(3) %>% equals(0) %>% ifelse("fizz",num),
         buzz=num %>% mod(5) %>% equals(0) %>% ifelse("buzz",fizz),
         fizzbuzz= num %>% mod(15) %>% equals(0) %>% ifelse("fizzbuzz",buzz)
         ) %>%
  select(num,fizzbuzz) %>%
  kable("markdown")

num	fizzbuzz
1	1
2	2
3	fizz
4	4
5	buzz
6	fizz
7	7
8	8
9	fizz
10	buzz
11	11
12	fizz
13	13
14	14
15	fizzbuzz
16	16
17	17
18	fizz
19	19
20	buzz
21	fizz
22	22
23	23
24	fizz
25	buzz
26	26
27	fizz
28	28
29	29
30	fizzbuzz
31	31
32	32
33	fizz
34	34
35	buzz
36	fizz
37	37
38	38
39	fizz
40	buzz
41	41
42	fizz
43	43
44	44
45	fizzbuzz
46	46
47	47
48	fizz
49	49
50	buzz
51	fizz
52	52
53	53
54	fizz
55	buzz
56	56
57	fizz
58	58
59	59
60	fizzbuzz
61	61
62	62
63	fizz
64	64
65	buzz
66	fizz
67	67
68	68
69	fizz
70	buzz
71	71
72	fizz
73	73
74	74
75	fizzbuzz
76	76
77	77
78	fizz
79	79
80	buzz
81	fizz
82	82
83	83
84	fizz
85	buzz
86	86
87	fizz
88	88
89	89
90	fizzbuzz
91	91
92	92
93	fizz
94	94
95	buzz
96	fizz
97	97
98	98
99	fizz
100	buzz

I hope nobody is solving this way on R interviews :)
Below easy to follow alternative:

f = seq(3,100,3)
b = seq(5,100,5)
fb = f[f %in% b] 
f = f[!f %in% fb]
b = b[!b %in% fb]
x = as.character(1:100)
x[f] = "Fizz"
x[b] = "Buzz"
x[fb] = "FizzBuzz"
cat(x, sep = "\n")