旋转更长的时间：R中的多行到列答案

【问题标题】：Pivot longer: Multiple rows to columns in R旋转更长的时间：R中的多行到列
【发布时间】：2020-06-12 11:49:15
【问题描述】：

我目前正在尝试研究如何旋转我的数据框（下面的小 dput）。目前一栏包含有关国家、ISO 代码、行业和部门的信息。我需要将此信息分散到 4 列中，并带有一个对应的值列。我之前使用过 melt 和 pivot_long 函数，但不确定如何生成 4 个新列以及 value 列。

DI_SMALL <- structure(list(V1 = structure(c(NA, NA, NA, NA, 1L, 1L, 1L, 1L
), .Label = "Energy Usage (TJ)", class = "factor"), V2 = structure(c(NA, 
NA, NA, NA, 2L, 1L, 4L, 3L), .Label = c("Coal", "Natural Gas", 
"Nuclear Electricity", "Petroleum"), class = "factor"), V3 = structure(c(5L, 
4L, 7L, 6L, 3L, 2L, 1L, 1L), .Label = c("0", "1.29327085460648e-05", 
"1.59504500372979e-05", "AFG", "Afghanistan", "Agriculture", 
"Industries"), class = "factor"), V4 = structure(c(5L, 4L, 7L, 
6L, 3L, 2L, 1L, 1L), .Label = c("0", "6.53466630114587e-06", 
"8.05944706428482e-06", "AFG", "Afghanistan", "Fishing", "Industries"
), class = "factor"), V5 = structure(c(5L, 4L, 6L, 7L, 3L, 2L, 
1L, 1L), .Label = c("0", "1.88562621206664e-05", "2.32557880912235e-05", 
"AFG", "Afghanistan", "Industries", "Mining and Quarrying"), class = "factor"), 
    V6 = structure(c(5L, 4L, 7L, 6L, 3L, 2L, 1L, 1L), .Label = c("0", 
    "2.00284547443433e-05", "2.47018365704401e-05", "AFG", "Afghanistan", 
    "Food & Beverages", "Industries"), class = "factor")), row.names = c("V1", 
"V2", "V3", "V4", "X", "X.1", "X.2", "X.3"), class = "data.frame")

理想情况下，输出将包含 7 列。现有的先到列，Country、ISO、Industry 和 Sector，然后是 Value。像这样：

Output <- structure(list(NA. = structure(c(1L, 1L, 1L, 1L), .Label = "Energy Usage (TJ)", class = "factor"), 
    NA..1 = structure(c(2L, 1L, 4L, 3L), .Label = c("Coal ", 
    "Natural Gas", "Nuclear Electricity", "Petroleum"), class = "factor"), 
    Country = structure(c(1L, 1L, 1L, 1L), .Label = "Afghanistan", class = "factor"), 
    ISO = structure(c(1L, 1L, 1L, 1L), .Label = "AFG", class = "factor"), 
    Industry = structure(c(1L, 1L, 1L, 1L), .Label = "Industries", class = "factor"), 
    Sector = structure(c(1L, 1L, 1L, 1L), .Label = "Agriculture", class = "factor"), 
    Value = c(1.595045004, 1.2932706, 0, 0)), class = "data.frame", row.names = c(NA, 
-4L))

希望这是有道理的，任何想法都将不胜感激！

谢谢

【问题讨论】：

在outputvalue是如何计算的？
您好，应该是DI_Small矩阵中的对应值，但都在一列中
以更易读的风格格式化您的代码会非常有帮助。例如。为每个参数创建一个新行。

标签： r pivot tidyverse tidyr reshape2

【解决方案1】：

使用data.table包，可以按照以下方式进行：

library(data.table)

setDT(DI_SMALL)[, V3 := as.character(V3)]
cols <- c("Country", "ISO", "Industry", "Sector")
Output <- DI_SMALL[, c(.(NA. = V1), .(NA..1 = V2), setNames(V3[is.na(V1)], cols), .(value = as.numeric(V3)))][!is.na(NA.)]

#                  NA.               NA..1     Country    ISO   Industry      Sector        value
# 1: Energy Usage (TJ)         Natural Gas Afghanistan    AFG Industries Agriculture 1.595045e-05
# 2: Energy Usage (TJ)                Coal Afghanistan    AFG Industries Agriculture 1.293271e-05
# 3: Energy Usage (TJ)           Petroleum Afghanistan    AFG Industries Agriculture 0.000000e+00
# 4: Energy Usage (TJ) Nuclear Electricity Afghanistan    AFG Industries Agriculture 0.000000e+00

【讨论】：

【解决方案2】：

在这里，您可以首先将 V3 重命名为 V6，并连接数据框的前 4 行，然后删除这些行，将数据框转换为更长的格式，最后通过拆分列“var”来创建所有四列" 使用包含新列名的重塑创建：

library(tidyr)
library(dplyr)
colNAMES <- apply(DI_SMALL[,3:6],2,function(x) paste(x[1:4],collapse="_"))
colnames(DI_SMALL)[3:6] <- colNAMES

DI_SMALL <- DI_SMALL[-c(1:4),]

DI_SMALL %>% pivot_longer(-c(V1,V2),names_to = "var",values_to = "Value") %>%
  mutate(Country = unlist(strsplit(var,"_"))[1],
         ISO = unlist(strsplit(var,"_"))[2],
         Industry = unlist(strsplit(var,"_"))[3],
         Sector = unlist(strsplit(var,"_"))[4]) %>%
  select(V1,V2,Country, ISO, Industry, Sector, Value)

# A tibble: 16 x 7
   V1                V2                  Country     ISO   Industry   Sector      Value               
   <fct>             <fct>               <chr>       <chr> <chr>      <chr>       <fct>               
 1 Energy Usage (TJ) Natural Gas         Afghanistan AFG   Industries Agriculture 1.59504500372979e-05
 2 Energy Usage (TJ) Natural Gas         Afghanistan AFG   Industries Agriculture 8.05944706428482e-06
 3 Energy Usage (TJ) Natural Gas         Afghanistan AFG   Industries Agriculture 2.32557880912235e-05
 4 Energy Usage (TJ) Natural Gas         Afghanistan AFG   Industries Agriculture 2.47018365704401e-05
 5 Energy Usage (TJ) Coal                Afghanistan AFG   Industries Agriculture 1.29327085460648e-05
 6 Energy Usage (TJ) Coal                Afghanistan AFG   Industries Agriculture 6.53466630114587e-06
 7 Energy Usage (TJ) Coal                Afghanistan AFG   Industries Agriculture 1.88562621206664e-05
 8 Energy Usage (TJ) Coal                Afghanistan AFG   Industries Agriculture 2.00284547443433e-05
 9 Energy Usage (TJ) Petroleum           Afghanistan AFG   Industries Agriculture 0                   
10 Energy Usage (TJ) Petroleum           Afghanistan AFG   Industries Agriculture 0                   
11 Energy Usage (TJ) Petroleum           Afghanistan AFG   Industries Agriculture 0                   
12 Energy Usage (TJ) Petroleum           Afghanistan AFG   Industries Agriculture 0                   
13 Energy Usage (TJ) Nuclear Electricity Afghanistan AFG   Industries Agriculture 0                   
14 Energy Usage (TJ) Nuclear Electricity Afghanistan AFG   Industries Agriculture 0                   
15 Energy Usage (TJ) Nuclear Electricity Afghanistan AFG   Industries Agriculture 0                   
16 Energy Usage (TJ) Nuclear Electricity Afghanistan AFG   Industries Agriculture 0

【讨论】：

【解决方案3】：

pivot_long 不适合这种情况，因为您将变量映射到行和列，并且它们不是列/行的名称。 Intead 您必须从变量中提取这些属性，然后“手动”构建data.frame。这是一个例子，我建议检查每个步骤中的变量值以便更好地理解这里的过程：

library(dplyr)

df <- DI_SMALL %>% 
  mutate_all(as.character) 

row_attr <-  paste0(df$V1, "/", df$V2)
row_attr <- row_attr[row_attr!= "NA/NA"]

col_attr <- df[1:4, -(1:2)] %>%
  apply(MARGIN = 2, function(x) paste0(x, collapse = "/"))

values <- df[-(1:4), -(1:2)] %>%
  mutate_all(as.numeric) %>%
  as.matrix() %>%
  c()

out <- expand.grid(row_attr, col_attr)
out <- cbind(out, values)

out <- out %>% 
  tidyr::separate(col = "Var1", into = c("NA.", "NA..1"), sep = "/") %>%
  tidyr::separate(col = "Var2", 
                  into = c("Country", "ISO", "Industry", "Sector"),
                  sep = "/")

out[1:4]

我认为Output 中的结果和DI_SMALL 的值具有不同的比例，但除此之外，这似乎是所需的输出。

                NA.               NA..1     Country ISO   Industry      Sector       values
1 Energy Usage (TJ)         Natural Gas Afghanistan AFG Industries Agriculture 1.595045e-05
2 Energy Usage (TJ)                Coal Afghanistan AFG Industries Agriculture 1.293271e-05
3 Energy Usage (TJ)           Petroleum Afghanistan AFG Industries Agriculture 0.000000e+00
4 Energy Usage (TJ) Nuclear Electricity Afghanistan AFG Industries Agriculture 0.000000e+00

【讨论】：

【解决方案4】：

我将首先对数据进行子集化，然后按如下方式从那里开始工作。尽管我仍然不确定您如何在所需的Output 中获得value。下面输出中的值与您在 MWE 中发布的内容不对应。希望这能给您带来领先优势。

subV<- as.data.frame(t(DI_SMALL[grep("V", rownames(DI_SMALL)), ]))[-c(1:2), ] # transpose `t()` this subset to get your desired variable levels into columns
subX<- DI_SMALL[grep("X", rownames(DI_SMALL)), 1:3]
Output <- cbind(subX[, 1:2],  subV, subX[, 3])
colnames(Output) <- c("NA.", "NA..1", "Country", "ISO", "Industry", "Sector", "Value"); rownames(Output) <- seq(1:nrow(Output))

> Output
                NA.               NA..1     Country ISO   Industry               Sector                Value
1 Energy Usage (TJ)         Natural Gas Afghanistan AFG Industries          Agriculture 1.59504500372979e-05
2 Energy Usage (TJ)                Coal Afghanistan AFG Industries              Fishing 1.29327085460648e-05
3 Energy Usage (TJ)           Petroleum Afghanistan AFG Industries Mining and Quarrying                    0
4 Energy Usage (TJ) Nuclear Electricity Afghanistan AFG Industries     Food & Beverages                    0

【讨论】：