﻿ Repeat Vector to Fill Down Column in Data Frame - ITCodar

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)]

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 "<-"