How do I repeat vector elements by times in another vector elements in R?

up vote
0
down vote

favorite

I want to repeat each

repeat.it <- c(0, 3951982, 7635488, 10986941)

by times:

repeat.times<- c(2L, 3L, 4L, 2L)

and get the result:

0,0,3951982,3951982,3951982,7635488,7635488,7635488,7635488,10986941,10986941

code I tried:
rep(repeat.it, each=repeat.times) but this seems to give me wrong results. How can I do it properly?

edited Nov 18 at 15:04

asked Nov 18 at 14:56

MAPK

1,607829

add a comment |

up vote
0
down vote

favorite

I want to repeat each

repeat.it <- c(0, 3951982, 7635488, 10986941)

by times:

repeat.times<- c(2L, 3L, 4L, 2L)

and get the result:

0,0,3951982,3951982,3951982,7635488,7635488,7635488,7635488,10986941,10986941

code I tried:
rep(repeat.it, each=repeat.times) but this seems to give me wrong results. How can I do it properly?

edited Nov 18 at 15:04

asked Nov 18 at 14:56

MAPK

1,607829

add a comment |

up vote
0
down vote

favorite

I want to repeat each

repeat.it <- c(0, 3951982, 7635488, 10986941)

by times:

repeat.times<- c(2L, 3L, 4L, 2L)

and get the result:

0,0,3951982,3951982,3951982,7635488,7635488,7635488,7635488,10986941,10986941

code I tried:
rep(repeat.it, each=repeat.times) but this seems to give me wrong results. How can I do it properly?

edited Nov 18 at 15:04

asked Nov 18 at 14:56

MAPK

1,607829

I want to repeat each

repeat.it <- c(0, 3951982, 7635488, 10986941)

by times:

repeat.times<- c(2L, 3L, 4L, 2L)

and get the result:

0,0,3951982,3951982,3951982,7635488,7635488,7635488,7635488,10986941,10986941

code I tried:
rep(repeat.it, each=repeat.times) but this seems to give me wrong results. How can I do it properly?

edited Nov 18 at 15:04

asked Nov 18 at 14:56

MAPK

1,607829

edited Nov 18 at 15:04

asked Nov 18 at 14:56

MAPK

1,607829

edited Nov 18 at 15:04

asked Nov 18 at 14:56

MAPK

1,607829

asked Nov 18 at 14:56

MAPK

1,607829

asked Nov 18 at 14:56

MAPK

1,607829

add a comment |

2 Answers
2

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oldest

votes

up vote
3
down vote

Almost there:

rep(repeat.it, times = repeat.times)

#  [1]        0        0  3951982  3951982  3951982  7635488  7635488

#  [8]  7635488  7635488 10986941 10986941

answered Nov 18 at 15:02

Julius Vainora

27.4k75878

@MAPK, does it answer your question?
– Julius Vainora
Nov 23 at 18:22

add a comment |

up vote
1
down vote

This is an option

res <- mapply(rep, repeat.it, repeat.times)

res <- unlist(res)

answered Nov 18 at 15:02

JRR

7891219

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
3
down vote

Almost there:

rep(repeat.it, times = repeat.times)

#  [1]        0        0  3951982  3951982  3951982  7635488  7635488

#  [8]  7635488  7635488 10986941 10986941

answered Nov 18 at 15:02

Julius Vainora

27.4k75878

@MAPK, does it answer your question?
– Julius Vainora
Nov 23 at 18:22

add a comment |

up vote
3
down vote

Almost there:

rep(repeat.it, times = repeat.times)

#  [1]        0        0  3951982  3951982  3951982  7635488  7635488

#  [8]  7635488  7635488 10986941 10986941

answered Nov 18 at 15:02

Julius Vainora

27.4k75878

@MAPK, does it answer your question?
– Julius Vainora
Nov 23 at 18:22

add a comment |

up vote
3
down vote

Almost there:

rep(repeat.it, times = repeat.times)

#  [1]        0        0  3951982  3951982  3951982  7635488  7635488

#  [8]  7635488  7635488 10986941 10986941

answered Nov 18 at 15:02

Julius Vainora

27.4k75878

Almost there:

rep(repeat.it, times = repeat.times)

#  [1]        0        0  3951982  3951982  3951982  7635488  7635488

#  [8]  7635488  7635488 10986941 10986941

answered Nov 18 at 15:02

Julius Vainora

27.4k75878

answered Nov 18 at 15:02

Julius Vainora

27.4k75878

answered Nov 18 at 15:02

Julius Vainora

27.4k75878

answered Nov 18 at 15:02

Julius Vainora

27.4k75878

@MAPK, does it answer your question?
– Julius Vainora
Nov 23 at 18:22

add a comment |

@MAPK, does it answer your question?
– Julius Vainora
Nov 23 at 18:22

@MAPK, does it answer your question?
– Julius Vainora
Nov 23 at 18:22

add a comment |

up vote
1
down vote

This is an option

res <- mapply(rep, repeat.it, repeat.times)

res <- unlist(res)

answered Nov 18 at 15:02

JRR

7891219

add a comment |

up vote
1
down vote

This is an option

res <- mapply(rep, repeat.it, repeat.times)

res <- unlist(res)

answered Nov 18 at 15:02

JRR

7891219

add a comment |

up vote
1
down vote

This is an option

res <- mapply(rep, repeat.it, repeat.times)

res <- unlist(res)

answered Nov 18 at 15:02

JRR

7891219

This is an option

res <- mapply(rep, repeat.it, repeat.times)

res <- unlist(res)

answered Nov 18 at 15:02

JRR

7891219

answered Nov 18 at 15:02

JRR

7891219

answered Nov 18 at 15:02

JRR

7891219

answered Nov 18 at 15:02

JRR

7891219

add a comment |

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