英文:
Generate new rows and assign values using data from specific parameters in dataframe
问题
我一直想使用特定参数为未知数据分配值。例如，我有特定湖泊在2004年的土地利用数据。然而，我没有这些湖泊在2002年和2003年的土地利用数据。有没有办法使用2004年的数据添加新的行，以包括2002-2003年的土地利用数据？我将提供一个展示我的数据现在是什么样子的数据集。
这是我想要的数据框的样子：
| 湖泊 | 年份 | 城市 | 农业 | 湿地 | 森林 | 灌木 |
|---|---|---|---|---|---|---|
| A | 2002 | 90 | 5 | 1 | 1 | 3 |
| A | 2003 | 90 | 5 | 1 | 1 | 3 |
| A | 2004 | 90 | 5 | 1 | 1 | 3 |
| B | 2002 | 80 | 1 | 4 | 5 | 10 |
| B | 2003 | 80 | 1 | 4 | 5 | 10 |
| B | 2004 | 80 | 1 | 4 | 5 | 10 |
| C | 2002 | 20 | 20 | 20 | 20 | 20 |
| C | 2003 | 20 | 20 | 20 | 20 | 20 |
| C | 2004 | 20 | 20 | 20 | 20 | 20 |
希望对您有所帮助。再次感谢！
英文:
I have been wanting to assign values for unknown data using specific parameters. For example, I have land usage data for specific lakes for the year 2004. However, I do not have land use data for these lakes for the years 2002 & 2003. Is there a way to add new rows with land use data for the years 2002-2003 using 2004 data? I will provide a data set for how my data looks like now.
| Lake | year | urban | agriculture | wetlands | forest | shrub |
|---|---|---|---|---|---|---|
| A | 2004 | 90 | 5 | 1 | 1 | 3 |
| B | 2004 | 80 | 1 | 4 | 5 | 10 |
| C | 2004 | 20 | 20 | 20 | 20 | 20 |
This is what I would like my data frame to look like:
| Lake | year | urban | agriculture | wetlands | forest | shrub |
|---|---|---|---|---|---|---|
| A | 2002 | 90 | 5 | 1 | 1 | 3 |
| A | 2003 | 90 | 5 | 1 | 1 | 3 |
| A | 2004 | 90 | 5 | 1 | 1 | 3 |
| B | 2002 | 80 | 1 | 4 | 5 | 10 |
| B | 2003 | 80 | 1 | 4 | 5 | 10 |
| B | 2004 | 80 | 1 | 4 | 5 | 10 |
| C | 2002 | 20 | 20 | 20 | 20 | 20 |
| C | 2003 | 20 | 20 | 20 | 20 | 20 |
| C | 2004 | 20 | 20 | 20 | 20 | 20 |
Any help is appreciated. Thanks again!
I do not know where to start so I have not tried anything.
答案1
得分: 1
毫无疑问，有很多方法可以做到这一点。如果这确实是您的用例，那么对于每一行，您可以将 year 转换为一个包含值 2002:2004 的列表。然后，只需使用 unnest() 函数对年份列表进行展开，所有其他数据将被复制以适用于所有 year 的值：
library(dplyr)library(tidyr)tab <- read.table(header=TRUE, textConnection("Lake year urban agriculture wetlands forest shrubA 2004 90 5 1 1 3B 2004 80 1 4 5 10C 2004 20 20 20 20 20"))tab %>%rowwise() %>%mutate(year = list(2002:2004)) %>%unnest(year)#> # A tibble: 9 × 7#> Lake year urban agriculture wetlands forest shrub#> <chr> <int> <int> <int> <int> <int> <int>#> 1 A 2002 90 5 1 1 3#> 2 A 2003 90 5 1 1 3#> 3 A 2004 90 5 1 1 3#> 4 B 2002 80 1 4 5 10#> 5 B 2003 80 1 4 5 10#> 6 B 2004 80 1 4 5 10#> 7 C 2002 20 20 20 20 20#> 8 C 2003 20 20 20 20 20#> 9 C 2004 20 20 20 20 20
创建于 2023-06-08，使用 reprex v2.0.2
英文:
No doubt there are lots of ways to do this. If this is really your use case, then for each row, you could turn year into a list with the values 2002:2004. Then, just unnest() the year list and all the other data will be replicated for all values of year:
library(dplyr)library(tidyr)tab <- read.table(header=TRUE, textConnection("Lake year urban agriculture wetlands forest shrubA 2004 90 5 1 1 3B 2004 80 1 4 5 10C 2004 20 20 20 20 20"))tab %>%rowwise() %>%mutate(year = list(2002:2004)) %>%unnest(year)#> # A tibble: 9 × 7#> Lake year urban agriculture wetlands forest shrub#> <chr> <int> <int> <int> <int> <int> <int>#> 1 A 2002 90 5 1 1 3#> 2 A 2003 90 5 1 1 3#> 3 A 2004 90 5 1 1 3#> 4 B 2002 80 1 4 5 10#> 5 B 2003 80 1 4 5 10#> 6 B 2004 80 1 4 5 10#> 7 C 2002 20 20 20 20 20#> 8 C 2003 20 20 20 20 20#> 9 C 2004 20 20 20 20 20
<sup>Created on 2023-06-08 with reprex v2.0.2</sup>
答案2
得分: 1
library(dplyr)library(tidyr)df |>uncount(3, .id = "year") |>mutate(year = year + 2001)
这段代码假设有一系列连续的3年，从2001年后开始（例如2002年，2003年，2004年）。
uncount将每一行重复3次，并添加一个名为year的列，该列仅是一个序列1:3，用于标识每个重复的行。mutate用于调整创建的year列，通过添加2001，将1:3转换为2002:2004。
输出
Lake year urban agriculture wetlands forest shrub1 A 2002 90 5 1 1 32 A 2003 90 5 1 1 33 A 2004 90 5 1 1 34 B 2002 80 1 4 5 105 B 2003 80 1 4 5 106 B 2004 80 1 4 5 107 C 2002 20 20 20 20 208 C 2003 20 20 20 20 209 C 2004 20 20 20 20 20
如果您想要非连续年份，可以像这样操作：
library(dplyr)library(tidyr)years <- c(2002, 2004) # 编辑这部分以获取您想要的年份df |>uncount(length(years)) |>mutate(year = rep(years, times = nrow(df)))
数据
df <- structure(list(Lake = c("A", "B", "C"), year = c(2004L, 2004L,2004L), urban = c(90L, 80L, 20L), agriculture = c(5L, 1L, 20L), wetlands = c(1L, 4L, 20L), forest = c(1L, 5L, 20L), shrub = c(3L,10L, 20L)), class = "data.frame", row.names = c(NA, -3L))
英文:
library(dplyr)library(tidyr)df |>uncount(3, .id = "year") |>mutate(year = year + 2001)
This assume a sequence of 3 consecutive years starting after 2001 (e.g. 2002, 2003, 2004).
uncountduplicates each row 3 times and adds the columnyearwhich is just a sequence1:3to identify each duplicated row.mutatewe use to adjust this createdyearby adding 2001, which transforms1:3into2002:2004.
Output
Lake year urban agriculture wetlands forest shrub1 A 2002 90 5 1 1 32 A 2003 90 5 1 1 33 A 2004 90 5 1 1 34 B 2002 80 1 4 5 105 B 2003 80 1 4 5 106 B 2004 80 1 4 5 107 C 2002 20 20 20 20 208 C 2003 20 20 20 20 209 C 2004 20 20 20 20 20
If you wanted to do non-sequential years you could do something like:
library(dplyr)library(tidyr)years <- c(2002, 2004) # edit this for years you wantdf |>uncount(length(years)) |>mutate(year = rep(years, times = nrow(df)))
Data
df <- structure(list(Lake = c("A", "B", "C"), year = c(2004L, 2004L,2004L), urban = c(90L, 80L, 20L), agriculture = c(5L, 1L, 20L), wetlands = c(1L, 4L, 20L), forest = c(1L, 5L, 20L), shrub = c(3L,10L, 20L)), class = "data.frame", row.names = c(NA, -3L))
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