Repeat Vector to Fill Down Column in Data Frame

Repeat vector to fill down column in data frame

If the vector can be evenly recycled, into the data.frame, you do not get and error or a warning:

df <- data.frame(x = 1:10)
df$z <- 1:5

This may be what you were experiencing before.

You can get your vector to fit as you mention with rep_len:

df$y <- rep_len(1:3, length.out=10)

This results in

df
    x z y
1   1 1 1
2   2 2 2
3   3 3 3
4   4 4 1
5   5 5 2
6   6 1 3
7   7 2 1
8   8 3 2
9   9 4 3
10 10 5 1

Note that in place of rep_len, you could use the more common rep function:

df$y <- rep(1:3,len=10)

From the help file for rep:

rep.int and rep_len are faster simplified versions for two common cases. They are not generic.

Mutate a column counting only 1 - 2 - 3 - 1 - 2 - 3

If number of rows is exactly divisible by the length of the vector then you can use :

df$row <- 1:3

A general solution would be to repeat 1:3 till number of rows in the dataframe.

df$row <- rep(1:3, length.out = nrow(df))

Fill data frame by column with for loop

Solution

L_df <- data.frame(sapply(pm, function(x) x * L * ((w - wu) / 100)))
names(L_df) <- c("L_por0", "L_por1", "L_por2", "L_por3", "L_por4", "L_por5",
                 "L_por6", "L_por7", "L_por8", "L_por9", "L_por10")
L_df
 L_por0    L_por1    L_por2    L_por3    L_por4    L_por5    L_por6    L_por7
1   1735.24  1717.888  1700.535  1683.183  1665.830  1648.478  1631.126  1613.773
2   3470.48  3435.775  3401.070  3366.366  3331.661  3296.956  3262.251  3227.546
3   5205.72  5153.663  5101.606  5049.548  4997.491  4945.434  4893.377  4841.320
4   6940.96  6871.550  6802.141  6732.731  6663.322  6593.912  6524.502  6455.093
5   8676.20  8589.438  8502.676  8415.914  8329.152  8242.390  8155.628  8068.866
6  10411.44 10307.326 10203.211 10099.097  9994.982  9890.868  9786.754  9682.639
7  12146.68 12025.213 11903.746 11782.280 11660.813 11539.346 11417.879 11296.412
8  13881.92 13743.101 13604.282 13465.462 13326.643 13187.824 13049.005 12910.186
9  15617.16 15460.988 15304.817 15148.645 14992.474 14836.302 14680.130 14523.959
10 17352.40 17178.876 17005.352 16831.828 16658.304 16484.780 16311.256 16137.732
11 19087.64 18896.764 18705.887 18515.011 18324.134 18133.258 17942.382 17751.505
12 20822.88 20614.651 20406.422 20198.194 19989.965 19781.736 19573.507 19365.278
13 22558.12 22332.539 22106.958 21881.376 21655.795 21430.214 21204.633 20979.052
14 24293.36 24050.426 23807.493 23564.559 23321.626 23078.692 22835.758 22592.825
15 26028.60 25768.314 25508.028 25247.742 24987.456 24727.170 24466.884 24206.598
      L_por8    L_por9   L_por10
1   1596.421  1579.068  1561.716
2   3192.842  3158.137  3123.432
3   4789.262  4737.205  4685.148
4   6385.683  6316.274  6246.864
5   7982.104  7895.342  7808.580
6   9578.525  9474.410  9370.296
7  11174.946 11053.479 10932.012
8  12771.366 12632.547 12493.728
9  14367.787 14211.616 14055.444
10 15964.208 15790.684 15617.160
11 17560.629 17369.752 17178.876
12 19157.050 18948.821 18740.592
13 20753.470 20527.889 20302.308
14 22349.891 22106.958 21864.024
15 23946.312 23686.026 23425.740

Explanation

The sapply() function can be used to iterate over vectors in a more idiomatic way for R programming. We iterate over pm and use your formula once since R is vectorised; each time it creates a vector of length 15 (so 11 vectors of length 15), and when we wrap it in data.frame() returns the data frame you want and we add in the column names.

NOTE: Applying functions to every element of a vector using an apply() family function has some different implications than iterating using for loops. In your case, I think sapply() is easier and more understandable. For more information on when you need a loop or when something like apply is better, see for example this discussion from Hadley Wickham's Advanced R book.

Repeat rows of a data.frame

df <- data.frame(a = 1:2, b = letters[1:2]) 
df[rep(seq_len(nrow(df)), each = 2), ]

Repeat rows of a data.frame N times

EDIT: updated to a better modern R answer.

You can use replicate(), then rbind the result back together. The rownames are automatically altered to run from 1:nrows.

d <- data.frame(a = c(1,2,3),b = c(1,2,3))
n <- 3
do.call("rbind", replicate(n, d, simplify = FALSE))

A more traditional way is to use indexing, but here the rowname altering is not quite so neat (but more informative):

 d[rep(seq_len(nrow(d)), n), ]

Here are improvements on the above, the first two using purrr functional programming, idiomatic purrr:

purrr::map_dfr(seq_len(3), ~d)

and less idiomatic purrr (identical result, though more awkward):

purrr::map_dfr(seq_len(3), function(x) d)

and finally via indexing rather than list apply using dplyr:

d %>% slice(rep(row_number(), 3))

Adding a column to data.frame and fill that column with a particular string using R

We can just assign the 'z'

df1$x1 <- 'z'

and then change the order of the columns,

df1[c(1, 6, 2:5)]

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Or if this column needs to be created in a particular position, we can use append

data.frame(append(df1, c(x1='z'), after=1))
#  y1 x1 x2 x3 x4 x5
#1  2  z  3  5  2  5
#2 11  z 13 34  4  4
#3 21  z  3 45 55 89

How to fill in rows with repeating data in pandas?

Seems there is no elegant way. This is the workaround I just figured out. Basically create a repeating list just bigger than original dataframe, and then left join them.

import pandas
df = pandas.DataFrame(range(100), columns=['first'])
repeat_arr = [1, 2, 3]
df = df.join(pandas.DataFrame(repeat_arr * (len(df)/len(repeat_arr)+1),
    columns=['second']))

Repeat Rows in Data Frame n Times

Use a combination of pd.DataFrame.loc and pd.Index.repeat

test.loc[test.index.repeat(test.times)]

  id  times
0  a      2
0  a      2
1  b      3
1  b      3
1  b      3
2  c      1
3  d      5
3  d      5
3  d      5
3  d      5
3  d      5

---

To mimic your exact output, use reset_index

test.loc[test.index.repeat(test.times)].reset_index(drop=True)

   id  times
0   a      2
1   a      2
2   b      3
3   b      3
4   b      3
5   c      1
6   d      5
7   d      5
8   d      5
9   d      5
10  d      5

How to use group_by() with rep_len() r

The dplyr package has a count function n() which could work.

mtcars %>%
     group_by(cyl) %>%
     mutate(newcol = rep_len(1:200, length.out=n()))

Also in the mutate statement it should be a "=" and not "<-"

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