英文:
How to use bind_rows for a list with incompatible sizes in dplyr?
问题
I have translated the code part as requested. Here is the translated code:
library(tidyverse)
df %>%
purrr::transpose() %>%
`$`(.borders) %>%
bind_rows() %>%
pivot_longer(everything(), values_to = "Borders")
Please let me know if you need any further assistance.
英文:
I am having a data like the following
$ECe
.type .value .borders .feature
1 ale -0.866 0.490 ECe
2 ale -0.811 2.680 ECe
3 ale -0.749 4.990 ECe
4 ale -0.501 7.994 ECe
5 ale -0.472 8.801 ECe
6 ale -0.364 11.140 ECe
7 ale -0.351 14.090 ECe
8 ale -0.172 15.900 ECe
9 ale -0.068 17.810 ECe
10 ale 0.069 18.750 ECe
11 ale 0.132 20.190 ECe
12 ale 0.176 21.020 ECe
13 ale 0.227 22.950 ECe
14 ale 0.295 24.520 ECe
15 ale 0.458 28.290 ECe
16 ale 0.534 31.640 ECe
17 ale 0.265 37.990 ECe
18 ale 0.330 40.700 ECe
19 ale 0.793 45.117 ECe
20 ale 0.393 53.090 ECe
21 ale 0.752 59.719 ECe
$BD
.type .value .borders .feature
1 ale 2.88401670 0.957 BD
2 ale 2.89757908 1.138 BD
3 ale 2.06229728 1.185 BD
4 ale 1.86256683 1.220 BD
5 ale 1.22004089 1.299 BD
6 ale 0.47244692 1.339 BD
7 ale 0.28531498 1.401 BD
8 ale 0.21050138 1.486 BD
9 ale 0.06942624 1.500 BD
10 ale -0.01446751 1.524 BD
11 ale -0.92647637 1.637 BD
12 ale -0.99773176 1.651 BD
13 ale -1.62164363 1.704 BD
14 ale -2.12120182 1.800 BD
15 ale -1.87179280 1.983 BD
16 ale -1.64798056 2.075 BD
17 ale -1.55269957 2.170 BD
18 ale -1.64049346 2.220 BD
I want to create a data frame like
# A tibble: 120 × 2
name Borders
<chr> <dbl>
1 ECe 0.491
2 SOC 0.261
3 BD 0.958
4 Sand 4.00
5 Silt 0
6 Clay 0
7 ECe 3.61
8 SOC 0.387
9 BD 1.02
10 Sand 8.42
# ℹ 110 more rows
I am using the following code
library(tidyverse)
df %>%
purrr::transpose() %>%
`$`(.borders) %>%
bind_rows() %>%
pivot_longer(everything(), values_to = "Borders")
But it returns the following error
>Error in recycle_columns():
! Tibble columns must have compatible sizes.
• Size 18: Columns BD and Clay.
• Size 20: Columns Sand and Silt.
• Size 21: Columns ECe and SOC.
ℹ Only values of size one are recycled.
How can I achieve this in R?
Here is the data in dput format
df <-
list(
ECe = structure(
list(
.type = c(
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale"
),
.value = c(
-0.866449860261058,-0.811874082709701,
-0.749993003099365,
-0.501212768536108,
-0.47268510237864,-0.364538974363897,
-0.351930853130633,
-0.172990179989361,
-0.0684651440322241,
0.0692627161333526,
0.132807737313057,
0.176914094922519,
0.227473041741352,
0.295586935903242,
0.458768756260148,
0.534728976733407,
0.265760631807008,
0.330347944699875,
0.793585231260766,
0.393724931630266,
0.752824545868792
),
.borders = c(
0.490806,
2.68,
4.99,
7.994,
8.801,
11.14,
14.09,
15.9,
17.81,
18.75,
20.19,
21.02,
22.95,
24.52,
28.29,
31.64,
37.99,
40.7,
45.1172,
53.09,
59.7192
),
.feature = c(
"ECe",
"ECe",
"ECe",
"ECe",
"ECe",
"ECe",
"ECe",
"ECe",
"ECe",
"ECe",
"ECe",
"ECe",
"ECe",
"ECe",
"ECe",
"ECe",
"ECe",
"ECe",
"ECe",
"ECe",
"ECe"
)
),
row.names = c(NA,-21L),
class = "data.frame"
),
SOC = structure(
list(
.type = c(
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale"
),
.value = c(
-1.12668195657973,-1.14070505975704,
-1.12096557264514,
-1.0364427676548,
-0.905420835797799,-0.683789433176536,
-0.53425868958447,
-0.483272349572241,-0.360467057150275,
-0.0571214846376487,
0.0130773857269451,
0.164166766380452,
0.233741319768938,
0.174784707378661,
0.188264622491949,
0.150568432447101,
0.532302724503294,
0.733931793684184,
1.1531117354405,
2.58573209158873,
2.75830864578748
),
.borders = c(
0.260869565217391,
0.330143540669856,
0.422727272727273,
0.545454545454545,
0.631578947368421,
0.686915887850467,
0.763636363636364,
0.802631578947368,
0.828571428571429,
0.889952153110048,
0.967289719626168,
1.02272727272727,
1.09090909090909,
1.23684210526316,
1.40520446096654,
1.52727272727273,
1.80861244019139,
1.90909090909091,
2.20093457943925,
2.49767441860465,
2.66511627906977
),
.feature = c(
"SOC",
"SOC",
"SOC",
"SOC",
"SOC",
"SOC",
"SOC",
"SOC",
"SOC",
"SOC",
"SOC",
"SOC",
"SOC",
"SOC",
"SOC",
"SOC",
"SOC",
"SOC",
"SOC",
"SOC",
"SOC"
)
),
row.names = c(NA,-21L),
class = "data.frame"
),
BD = structure(
list(
.type = c(
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale"
),
.value = c(
2.88401670431712,
2.89757908472024,
2.06229728227612,
1.8625668296819,
1.22004089065561,
0.472446924052171,
0.285314983167609,
0.210501378242261,
0.0694262373804251,-0.0144675088601809,
-0.926476372785716,
-0.997731763217841,-1.62164362525624,
-2.12120181627754,
-1.87179280480865,
-1.64798056454827,-1.5526995652836,
-1.64049345720051
),
.borders = c(
0.957710151765353,
1.13854777070064,
1.18522292993631,
1.22058598726115,
1.29930014940631,
1.33971974522293,
1.40121097743178,
1.48638829912715,
1.5,
1.52428025477707,
1.6371974522293,
1.65183612487222,
1.70405095541401,
1.80025477707006,
1.98361248722183,
2.07583549579303,
2.17044900526854,
2.22039789258473
),
.feature = c(
"BD",
"BD",
"BD",
"BD",
"BD",
"BD",
"BD",
"BD",
"BD",
"BD",
"BD",
"BD",
"BD",
"BD",
"BD",
"BD",
"BD",
"BD"
)
),
row.names = c(NA,-18L),
class = "data.frame"
),
Sand = structure(
list(
.type = c(
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale"
),
.value = c(
5.72243888922004,
5.04938965197184,
4.7017193997508,
4.38186986390524,
3.55049136618825,
3.79907192663079,
3.65192607277815,
2.69221076986002,
2.08417035244566,
0.508134498143654,
-0.760157331236183,-2.22196153094718,
-2.60183427497522,
-2.6090089814882,
-4.29829842975557,-4.89113681989785,
-6.17719615317024,
-6.51041335854504,
-6.70721878970499,-6.69985492661679
),
.borders = c(
3.9999999999992,
7.99999999999983,
12.0000000000005,
20.0000000000003,
23.9999999999995,
27.9999999999987,
31.9999999999993,
39.9999999999991,
43.9999999999998,
48.0000000000004,
52.000000000001,
59.9999999999994,
63.9999999999986,
64.0000000000001,
68.0000000000007,
71.9999999999999,
76.0000000000005,
79.9999999999997,
84.0000000000003,
88.000000000001
),
.feature = c(
"Sand",
"Sand",
"Sand",
"Sand",
"Sand",
"Sand",
"Sand",
"Sand",
"Sand",
"Sand",
"Sand",
"Sand",
"Sand",
"Sand",
"Sand",
"Sand",
"Sand",
"Sand",
"Sand",
"Sand"
)
),
row.names = c(NA,-20L),
class = "data.frame"
),
Silt = structure(
list(
.type = c(
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale"
),
.value = c(
-1.81467365524819,-1.80595765781436,
-1.44252832794551,
-1.40678289959526,-1.3809409627197,
-1.27631650711468,
-0.918535212874314,-0.891479949841867,
-0.130663280250859,
0.054226944600215,
0.0371390649359506,
0.547567627523539,
0.540716330465325,
1.14691203242341,
1.45243650519454,
1.51824963792443,
1.76292774269082,
1.85725359158924,
2.48581029669555,
0.388409658807059
),
.borders = c(
0,
3.99999999999921,
7.99999999999841,
7.99999999999983,
8.00000000000125,
11.999999999999,
15.9999999999997,
16.0000000000011,
20.0000000000003,
23.9999999999995,
24.0000000000009,
27.9999999999987,
28.0000000000001,
31.9999999999993,
35.9999999999985,
40.0000000000006,
43.9999999999998,
47.999999999999,
64.0000000000015,
79.9999999999997
),
.feature = c(
"Silt",
"Silt",
"Silt",
"Silt",
"Silt",
"Silt",
"Silt",
"Silt",
"Silt",
"Silt",
"Silt",
"Silt",
"Silt",
"Silt",
"Silt",
"Silt",
"Silt",
"Silt",
"Silt",
"Silt"
)
),
row.names = c(NA,-20L),
class = "data.frame"
),
Clay = structure(
list(
.type = c(
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale",
"ale"
),
.value = c(
-1.02199413478959,-0.826304780697195,
-0.942406575344331,
-0.971009582383038,-1.24981505440423,
-1.24219672780758,
-0.443954661148336,-0.344646636267013,
-0.344646636267013,
0.0190677799370838,
0.509523035852588,
0.805292224788913,
0.990611614682741,
0.998147256094054,
1.16179870265779,
2.06234942853985,
2.8900928986798,
3.82397972588229
),
.borders = c(
0,
7.99999999999983,
11.999999999999,
12.0000000000005,
15.9999999999997,
16.0000000000011,
20.0000000000003,
23.9999999999995,
24.0000000000009,
28.0000000000001,
31.9999999999993,
35.9999999999999,
39.9999999999991,
40.0000000000006,
47.999999999999,
51.9999999999996,
56.0000000000002,
67.9999999999993
),
.feature = c(
"Clay",
"Clay",
"Clay",
"Clay",
"Clay",
"Clay",
"Clay",
"Clay",
"Clay",
"Clay",
"Clay",
"Clay",
"Clay",
"Clay",
"Clay",
"Clay",
"Clay",
"Clay"
)
),
row.names = c(NA,-18L),
class = "data.frame"
)
)
答案1
得分: 2
以下是翻译好的内容：
使用 map() 和 list_rbind()：
library(tidyverse)
map(df, ~ select(.x, Name = .feature, Borders = .borders, Value = .value)) %>%
list_rbind() %>%
group_by(Name) %>%
mutate(row_order = 1:n()) %>%
ungroup() %>%
arrange(row_order) %>%
select(-row_order)
# 一个数据框：118 行 × 3 列
Name Borders Value
<chr> <dbl> <dbl>
1 ECe 0.491 -0.866
2 SOC 0.261 -1.13
3 BD 0.958 2.88
4 Sand 4.00 5.72
5 Silt 0 -1.81
6 Clay 0 -1.02
7 ECe 2.68 -0.812
8 SOC 0.330 -1.14
9 BD 1.14 2.90
10 Sand 8.00 5.05
# … 还有 108 行
如果顺序不重要，您可以使用以下方式：
map(df, ~ select(.x, Name = .feature, Borders = .borders, Value = .value)) %>%
list_rbind()
英文:
Use map() and list_rbind():
library(tidyverse)
map(df, ~ select(.x, Name = .feature, Borders = .borders, Value = .value)) |>
list_rbind() |>
group_by(Name) |>
mutate(row_order = 1:n()) |>
ungroup() |>
arrange(row_order) |>
select(-row_order)
# A tibble: 118 × 3
Name Borders Value
<chr> <dbl> <dbl>
1 ECe 0.491 -0.866
2 SOC 0.261 -1.13
3 BD 0.958 2.88
4 Sand 4.00 5.72
5 Silt 0 -1.81
6 Clay 0 -1.02
7 ECe 2.68 -0.812
8 SOC 0.330 -1.14
9 BD 1.14 2.90
10 Sand 8.00 5.05
# … with 108 more rows
If the order does not matter, you can just go with:
map(df, ~ select(.x, Name = .feature, Borders = .borders, Value = .value)) |>
list_rbind()
答案2
得分: 2
Here is the translated content you requested:
Udpate OP请求见评论：
要添加.value列：
imap_dfr(df, ~ data.frame(name = .y, Borders = .x$.borders, value = .x$.value))
name Borders value
<chr> <dbl> <dbl>
1 ECe 0.491 -0.866
2 ECe 2.68 -0.812
3 ECe 4.99 -0.750
4 ECe 7.99 -0.501
5 ECe 8.80 -0.473
6 ECe 11.1 -0.365
7 ECe 14.1 -0.352
8 ECe 15.9 -0.173
9 ECe 17.8 -0.0685
10 ECe 18.8 0.0693
# … with 108 more rows
第一个答案：
为了解决您遇到的错误：
- 使用
purrr::transpose() %>%$(.borders)生成了不同长度的6个向量的列表。
然后您尝试将它们按列与bind_rows组合到一个tibble中。错误消息显示：
• Size 18: Columns `BD` and `Clay`.
• Size 20: Columns `Sand` and `Silt`.
• Size 21: Columns `ECe` and `SOC`.
ℹ Only values of size one are recycled.```
要实现您想要的操作，您可以使用 `purrr::map_df()` 或其变体 `purrr::imap_dfr()`：
这是使用 `map_df` 的版本：
```R
library(purrr)
library(dplyr)
map(df, ~(.x %>% select(c(.feature, .borders)))) %>%
map_df(., ~ .x %>% select(.borders), .id = "name")
或者
library(purrr)
library(dplyr)
imap_dfr(df, ~ data.frame(name = .y, Borders = .x$.borders))
name .borders
1 ECe 0.4908060
2 ECe 2.6800000
3 ECe 4.9900000
4 ECe 7.9940000
5 ECe 8.8010000
6 ECe 11.1400000
7 ECe 14.0900000
8 ECe 15.9000000
9 ECe 17.8100000
10 ECe 18.7500000
11 ECe 20.1900000
12 ECe 21.0200000
13 ECe 22.9500000
14 ECe 24.5200000
15 ECe 28.2900000
16 ECe 31.6400000
17 ECe 37.9900000
18 ECe 40.7000000
19 ECe 45.1172000
20 ECe 53.0900000
21 ECe 59.7192000
22 SOC 0.2608696
23 SOC 0.3301435
24 SOC 0.4227273
25 SOC 0.5454545
26 SOC 0.6315789
27 SOC 0.6869159
28 SOC 0.7636364
29 SOC 0.8026316
30 SOC 0.8285714
31 SOC 0.8899522
32 SOC 0.9672897
33 SOC 1.0227273
34 SOC 1.0909091
35 SOC 1.2368421
36 SOC 1.4052045
37 SOC 1.5272727
38 SOC 1.8086124
39 SOC 1.9090909
40 SOC 2.2009346
41 SOC 2.4976744
42 SOC 2.6651163
43 BD 0.9577102
44 BD 1.1385478
45 BD 1.1852229
46 BD 1.2205860
47 BD 1.2993001
48 BD 1.3397197
49 BD 1.4012110
50 BD 1.4863883
51 BD 1.5000000
52 BD 1.5242803
53 BD 1.6371975
54 BD 1.6518361
55 BD 1.7040510
56 BD 1.8002548
57 BD 1.9836125
58 BD 2.0758355
59 BD 2.1704490
60 BD 2.2203979
61 Sand 4.0000000
62 Sand 8.0000000
63 Sand 12.0000000
64 Sand 20.0000000
65 Sand 24.0000000
66 Sand 28.0000000
67 Sand 32.0000000
68 Sand 40.0000000
69 Sand 44.0000000
70 Sand 48.0000000
71 Sand 52.0000000
72 Sand 60.0000000
73 Sand 64.0000000
74 Sand 64.0000000
75 Sand 68.0000000
76 Sand 72.0000000
77 Sand 76.0000000
78 Sand 80.0000000
79 Sand 84.0000000
80 Sand 88.0000000
81 Silt 0.0000000
82 Silt 4.0000000
83 Silt 8.0000000
84 Silt 8.0000000
85 Silt 8.0000000
86 Silt 12.0000000
87 Silt 16.0000000
88 Silt 16.0000000
89 Silt 20.0000000
90 Silt 24.0000000
91 Silt 24.0000000
92 Silt 28.0000000
93 Silt 28.000000
<details>
<summary>英文:</summary>
**Udpate** OP request see comments:
to add .value column:
imap_dfr(df, ~ data.frame(name = .y, Borders = .x$.borders, value = .x$.value))
name Borders value
<chr> <dbl> <dbl>
1 ECe 0.491 -0.866
2 ECe 2.68 -0.812
3 ECe 4.99 -0.750
4 ECe 7.99 -0.501
5 ECe 8.80 -0.473
6 ECe 11.1 -0.365
7 ECe 14.1 -0.352
8 ECe 15.9 -0.173
9 ECe 17.8 -0.0685
10 ECe 18.8 0.0693
… with 108 more rows
**First answer:**
To address the error you get:
1. With `purrr::transpose() %>%
`$`(.borders)` you create list of 6 vectors of different length.
And then you try to combine them columnwise with bind_rows to a tibble. And the error says
``` ! Tibble columns must have compatible sizes.
• Size 18: Columns `BD` and `Clay`.
• Size 20: Columns `Sand` and `Silt`.
• Size 21: Columns `ECe` and `SOC`.
ℹ Only values of size one are recycled.```
To get what you want to do, you could either use purrr::map_df() or its variant purrr::imap_dfr():
Here is a version using map_df:
library(purrr)
library(dplyr)
map(df, ~(.x %>% select(c(.feature, .borders)))) %>%
map_df(., ~ .x %>% select(.borders), .id = "name")
OR
library(purrr)
library(dplyr)
imap_dfr(df, ~ data.frame(name = .y, Borders = .x$.borders))
name .borders
1 ECe 0.4908060
2 ECe 2.6800000
3 ECe 4.9900000
4 ECe 7.9940000
5 ECe 8.8010000
6 ECe 11.1400000
7 ECe 14.0900000
8 ECe 15.9000000
9 ECe 17.8100000
10 ECe 18.7500000
11 ECe 20.1900000
12 ECe 21.0200000
13 ECe 22.9500000
14 ECe 24.5200000
15 ECe 28.2900000
16 ECe 31.6400000
17 ECe 37.9900000
18 ECe 40.7000000
19 ECe 45.1172000
20 ECe 53.0900000
21 ECe 59.7192000
22 SOC 0.2608696
23 SOC 0.3301435
24 SOC 0.4227273
25 SOC 0.5454545
26 SOC 0.6315789
27 SOC 0.6869159
28 SOC 0.7636364
29 SOC 0.8026316
30 SOC 0.8285714
31 SOC 0.8899522
32 SOC 0.9672897
33 SOC 1.0227273
34 SOC 1.0909091
35 SOC 1.2368421
36 SOC 1.4052045
37 SOC 1.5272727
38 SOC 1.8086124
39 SOC 1.9090909
40 SOC 2.2009346
41 SOC 2.4976744
42 SOC 2.6651163
43 BD 0.9577102
44 BD 1.1385478
45 BD 1.1852229
46 BD 1.2205860
47 BD 1.2993001
48 BD 1.3397197
49 BD 1.4012110
50 BD 1.4863883
51 BD 1.5000000
52 BD 1.5242803
53 BD 1.6371975
54 BD 1.6518361
55 BD 1.7040510
56 BD 1.8002548
57 BD 1.9836125
58 BD 2.0758355
59 BD 2.1704490
60 BD 2.2203979
61 Sand 4.0000000
62 Sand 8.0000000
63 Sand 12.0000000
64 Sand 20.0000000
65 Sand 24.0000000
66 Sand 28.0000000
67 Sand 32.0000000
68 Sand 40.0000000
69 Sand 44.0000000
70 Sand 48.0000000
71 Sand 52.0000000
72 Sand 60.0000000
73 Sand 64.0000000
74 Sand 64.0000000
75 Sand 68.0000000
76 Sand 72.0000000
77 Sand 76.0000000
78 Sand 80.0000000
79 Sand 84.0000000
80 Sand 88.0000000
81 Silt 0.0000000
82 Silt 4.0000000
83 Silt 8.0000000
84 Silt 8.0000000
85 Silt 8.0000000
86 Silt 12.0000000
87 Silt 16.0000000
88 Silt 16.0000000
89 Silt 20.0000000
90 Silt 24.0000000
91 Silt 24.0000000
92 Silt 28.0000000
93 Silt 28.0000000
94 Silt 32.0000000
95 Silt 36.0000000
96 Silt 40.0000000
97 Silt 44.0000000
98 Silt 48.0000000
99 Silt 64.0000000
100 Silt 80.0000000
101 Clay 0.0000000
102 Clay 8.0000000
103 Clay 12.0000000
104 Clay 12.0000000
105 Clay 16.0000000
106 Clay 16.0000000
107 Clay 20.0000000
108 Clay 24.0000000
109 Clay 24.0000000
110 Clay 28.0000000
111 Clay 32.0000000
112 Clay 36.0000000
113 Clay 40.0000000
114 Clay 40.0000000
115 Clay 48.0000000
116 Clay 52.0000000
117 Clay 56.0000000
118 Clay 68.0000000
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
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