How to Add a Row to a Data Frame in R

How to add a row to a data frame in R?

Like @Khashaa and @Richard Scriven point out in comments, you have to set consistent column names for all the data frames you want to append.

Hence, you need to explicitly declare the columns names for the second data frame, de, then use rbind(). You only set column names for the first data frame, df:

df<-data.frame("hi","bye")
names(df)<-c("hello","goodbye")

de<-data.frame("hola","ciao")
names(de)<-c("hello","goodbye")

newdf <- rbind(df, de)

Adding named rows to a data.frame in R

You can use rbind() on the named vectors.

rbind(d, com = 0:3, min = 2:5) 

    ESL prof scope type
A     1    0     2    3
B     2    1     3    4
C     3    2     4    5
D     4    3     5    6
E     5    4     6    7
com   0    1     2    3
min   2    3     4    5

Add a row with calculations in R

Does this work:

library(dplyr)
library(stringr)
x %>% 
  bind_rows(x %>% summarise(a = str_c(a, lead(a), sep = '_')) %>% na.omit() %>% 
                bind_cols(x %>% summarise(across(2:4, ~ . - lead(.))) %>% na.omit()))
    a ja fe ma
1   I 30 50 35
2   E 20 40 22
3 I_E 10 10 13

How add a row of 0 to a dataframe

We can use rbind

mat <- rbind(0, mat)

-output

mat
   a  b  c  d  e
1  0  0  0  0  0
2  2 45 77 52 14
3  5 78 85 68 73
4 90 98  3  4 91
5 77 55 22 25 87
6 56 63  4 79 94

Add rows to data frame by a pattern

This may be an alternative approach:

library(tidyverse)

df1 %>%
  mutate(rn = row_number()) %>%
  pivot_wider(id_cols = c(ID, rn), names_from = Order, values_from = Value) %>%
  mutate(post2 = if_else(!is.na(lead(post)), lead(post), pre),
         pre2 = if_else(!is.na(post2) & is.na(pre), post2, pre)) %>%
  select(-c(rn, pre, post)) %>%
  pivot_longer(cols = c(general, pre2, post2), names_to = "Order", values_to = "Value") %>%
  drop_na()

Output

# A tibble: 19 x 3
   ID    Order   Value
   <fct> <chr>   <dbl>
 1 1     general     1
 2 1     pre2        3
 3 1     post2       4
 4 1     pre2        7
 5 1     post2       7
 6 1     pre2        0
 7 1     post2      10
 8 2     general     1
 9 2     pre2        0
10 2     post2       0
11 2     pre2       12
12 2     post2      12
13 3     general    12
14 3     pre2        3
15 3     post2       4
16 3     pre2        6
17 3     post2       6
18 3     pre2        8
19 3     post2       8

Edit:

To generalize this solution for multiple Value columns, you will need to first pivot_longer to put data into a more workable format. In addition, you will want to group_by the column name variable so that using lead you are only looking at values appropriate for that variable.

Say for example you have two columns, Value1 and Value2:

df1 <- data.frame("ID" = as.factor(c('1', '1', '1', '1', '1', '1', '2', '2', '2', '3', '3', '3', '3', '3')), 
                  "Order" = as.factor(c('general', 'pre', 'post', 'post', 'pre', 'post', 'general', 'post', 'pre', 'general', 'pre', 'post', 'pre', 'pre')), 
                  "Value1" = as.numeric(c('1', '3','4','7','0','10', '1','0','12', '12', '3', '4', '6','8')),
                  "Value2" = as.numeric(c('4', '2','1','9','2','15', '2','11','18', '16', '5', '5', '8','10')))

You can do the following:

df1 %>%
  pivot_longer(cols = starts_with("Value"), names_to = "ValueName", values_to = "Value") %>%
  mutate(rn = row_number()) %>%
  pivot_wider(id_cols = c(ID, rn, ValueName), names_from = Order, values_from = Value) %>%
  group_by(ID, ValueName) %>%
  mutate(post2 = if_else(!is.na(lead(post)), lead(post), pre),
         pre2 = if_else(!is.na(post2) & is.na(pre), post2, pre)) %>%
  select(-c(rn, pre, post)) %>%
  rename(pre = pre2, post = post2) %>%
  pivot_longer(cols = c(general, pre, post), names_to = "Order", values_to = "Value") %>%
  drop_na() %>%
  arrange(ValueName, ID) %>%
  print(n=50)

Output

# A tibble: 38 x 4
# Groups:   ID, ValueName [6]
   ID    ValueName Order   Value
   <fct> <chr>     <chr>   <dbl>
 1 1     Value1    general     1
 2 1     Value1    pre         3
 3 1     Value1    post        4
 4 1     Value1    pre         7
 5 1     Value1    post        7
 6 1     Value1    pre         0
 7 1     Value1    post       10
 8 2     Value1    general     1
 9 2     Value1    pre         0
10 2     Value1    post        0
11 2     Value1    pre        12
12 2     Value1    post       12
13 3     Value1    general    12
14 3     Value1    pre         3
15 3     Value1    post        4
16 3     Value1    pre         6
17 3     Value1    post        6
18 3     Value1    pre         8
19 3     Value1    post        8
20 1     Value2    general     4
21 1     Value2    pre         2
22 1     Value2    post        1
23 1     Value2    pre         9
24 1     Value2    post        9
25 1     Value2    pre         2
26 1     Value2    post       15
27 2     Value2    general     2
28 2     Value2    pre        11
29 2     Value2    post       11
30 2     Value2    pre        18
31 2     Value2    post       18
32 3     Value2    general    16
33 3     Value2    pre         5
34 3     Value2    post        5
35 3     Value2    pre         8
36 3     Value2    post        8
37 3     Value2    pre        10
38 3     Value2    post       10

The data is left in long format - but could be converted to wide as well in the end with pivot_wider.

Adding row of one data frame at a specific spot in another dataframe in R

I hope I got the logic right for your problem ... I did it for a data.frame of 30 rows to add a row every 10 rows (as 120 is to much for a reproducable example in terms of fitting the output in the answer).

library(dplyr)

r <- 3 # your number is 46 (5520/120)
l <- 10 # your number is 120

# your long data.frame where you want to fit in ever l rows
df1 <- data.frame(dx = c("a","a","a","a","a","a","a","a","a","a",
                         "c","c","c","c","c","c","c","c","c","c", 
                         "e","e","e","e","e","e","e","e","e","e"))

# your data.frame of one row to fit in every l rows
df2 <- data.frame(dy = c("X"))

# set colnames to be identical
names(df2) <- colnames(df1) 

# use row number as ID and set it of as needed with the help of integer division
dff1 <- df1 %>% 
  dplyr::mutate(ID = dplyr::row_number()) %>% 
  dplyr::mutate(ID = ID + (ID-1) %/% l)

# repeat your one row df according to the quantity needed and use the row number with set off calculation
dff2 <- df2 %>% 
  dplyr::slice(rep(row_number(), r)) %>% 
  dplyr::mutate(ID = dplyr::row_number()) %>% 
  dplyr::mutate(ID = (ID) * l + ID)

# union both data.frames (I am supposing column types are identical!)
dff1 %>% 
  dplyr::union(dff2) %>% 
  dplyr::arrange(ID)

   dx ID
1   a  1
2   a  2
3   a  3
4   a  4
5   a  5
6   a  6
7   a  7
8   a  8
9   a  9
10  a 10
11  X 11
12  c 12
13  c 13
14  c 14
15  c 15
16  c 16
17  c 17
18  c 18
19  c 19
20  c 20
21  c 21
22  X 22
23  e 23
24  e 24
25  e 25
26  e 26
27  e 27
28  e 28
29  e 29
30  e 30
31  e 31
32  e 32
33  X 33

Add a value as a new row to a dataframe but keep all other columns NA

A possible solution, based on dplyr. We first need to convert ID from numeric to character.

library(dplyr)

mydata %>% 
  mutate(ID = as.character(ID)) %>% 
  bind_rows(list(ID = "School 1"), .)

#> # A tibble: 6 × 3
#>   ID       subject   age
#>   <chr>    <chr>   <dbl>
#> 1 School 1 <NA>       NA
#> 2 1        His        21
#> 3 2        Geo        24
#> 4 3        Geo        26
#> 5 4        His        23
#> 6 5        Geo        26

How to Add a Row to a Data Frame in R