CODEWITHSUNDEEP
rlistdplyr
Devin
Devin

Reputation: 323

Bind rows by by select column names from a list?

I have a list of date frames:

x <- data.frame("SN" = 1:25,"Age" = 1:25,"Gender" = rep(c("Male","Female"),25))
y <- data.frame("SN" = 1:25,"Age" = 1:25,"Occupation" = rep(c("Barber","Doctor"),25))
z <- data.frame("SN" = 1:25,"Age" = 1:25)
list <- c(x,y,z)

I am attempting to bind the data frames together into one data frame, unlisted. I want to pull only specific columns to make the end data frame however.

So for instance, I want x,y, and z combined into one data frame with only SN and Age in the resulting data frame.

Is there a simple way of doing this?

Upvotes: 1

Views: 2452

Answers (8)

KWigg
KWigg

Reputation: 16

I approached it the same way as igorkf but set my syntax slightly differently:

xyz <- x %>% select(SN, age) %>%      
    bind_rows(y %>% select(SN, age) %>%       
    bind_rows(z %>% select(SN, age))

Upvotes: 0

Chris G
Chris G

Reputation: 64

Too simple? Are Age and SB always in the 1 and 2 spot?

list <- as.data.frame(c(x[1:2],y[1:2],z[1:2]))

Upvotes: 1

ericOss
ericOss

Reputation: 181

Is this simple enough?

xyz <- bind_rows(select(x, SN, Age), select(y, SN, Age), select(z, SN, Age))

Upvotes: 3

JasonAizkalns
JasonAizkalns

Reputation: 20463

An approach with purrr:

library(purrr)
lst %>% map_dfr(`[`, c("SN", "Age"))

Which says, map the extract [ function over the items "SN" and "Age" from each list, then bind together all those elements into a data.frame df -- and hey, while you're binding them together, please bind the rows so dfr.

Upvotes: 4

markus
markus

Reputation: 26343

Given 

lst <- c(x,y,z)

it turns out that we can do

out <- unstack(stack(lst))
head(out)
#  SN Age
#1  1   1
#2  2   2
#3  3   3
#4  4   4
#5  5   5
#6  6   6

Upvotes: 2

Prevost
Prevost

Reputation: 697

Here's a data.table solution that will work with any data.table that has any number of columns. It lines up the columns based on their names, and fills any blanks with NA where needed. You will can always use that same approach for any number of data.tables.

library(data.table)
library(magrittr)

x <- data.frame("SN" = 1:25,"Age" = 1:25,"Gender" = rep(c("Male","Female"),25))
y <- data.frame("SN" = 1:25,"Age" = 1:25,"Occupation" = rep(c("Barber","Doctor"),25))
z <- data.frame("SN" = 1:25,"Age" = 1:25)

listy <- list(x,y,z)

a <- rbindlist(listy,
               use.names = TRUE,
               fill = TRUE) %>% 
  .[, .(SN, Age)]

Created on 2019-02-27 by the reprex package (v0.2.1)

Upvotes: 2

fujiu
fujiu

Reputation: 501

In case you want a general solution that will also work when you have more than three elements in your list of dataframes:

library(dplyr)

x <- data.frame("SN" = 1:25,"Age" = 1:25,"Gender" = rep(c("Male","Female"),25))
y <- data.frame("SN" = 1:25,"Age" = 1:25,"Occupation" = rep(c("Barber","Doctor"),25))
z <- data.frame("SN" = 1:25,"Age" = 1:25)
lst <- list(x,y,z)

df <- do.call(rbind, lapply(lst, select, SN, Age))

Upvotes: 4

igorkf
igorkf

Reputation: 3565

I don't got your question, but this helps?

bind_rows( (x %>% select(SN, Age)), (y %>% select(SN, Age)), z)
    SN Age
1    1   1
2    2   2
3    3   3
4    4   4
5    5   5
6    6   6
7    7   7
8    8   8
9    9   9
10  10  10
11  11  11
12  12  12
13  13  13
14  14  14
15  15  15
16  16  16
17  17  17
18  18  18
19  19  19
20  20  20
21  21  21
22  22  22
23  23  23
24  24  24
25  25  25
26   1   1
27   2   2
28   3   3
29   4   4
30   5   5
31   6   6
32   7   7
33   8   8
34   9   9
35  10  10
36  11  11
37  12  12
38  13  13
39  14  14
40  15  15
41  16  16
42  17  17
43  18  18
44  19  19
45  20  20
46  21  21
47  22  22
48  23  23
49  24  24
50  25  25
51   1   1
52   2   2
53   3   3
54   4   4
55   5   5
56   6   6
57   7   7
58   8   8
59   9   9
60  10  10
61  11  11
62  12  12
63  13  13
64  14  14
65  15  15
66  16  16
67  17  17
68  18  18
69  19  19
70  20  20
71  21  21
72  22  22
73  23  23
74  24  24
75  25  25
76   1   1
77   2   2
78   3   3
79   4   4
80   5   5
81   6   6
82   7   7
83   8   8
84   9   9
85  10  10
86  11  11
87  12  12
88  13  13
89  14  14
90  15  15
91  16  16
92  17  17
93  18  18
94  19  19
95  20  20
96  21  21
97  22  22
98  23  23
99  24  24
100 25  25
101  1   1
102  2   2
103  3   3
104  4   4
105  5   5
106  6   6
107  7   7
108  8   8
109  9   9
110 10  10
111 11  11
112 12  12
113 13  13
114 14  14
115 15  15
116 16  16
117 17  17
118 18  18
119 19  19
120 20  20
121 21  21
122 22  22
123 23  23
124 24  24
125 25  25

Upvotes: 2

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