我想用统计列Innovation分散第一列min max mean sd。
这是我开发的代码：
plot1 <- BANKRUPTCY2 %>%
select(Innovation,DelTradMerge,Depth,Scope,Novelty,PatW,Assets,Age,Sales,Income,NasdaqIndex) %>%
group_by(Innovation,DelTradMerge) %>%
rename_all(. %>% gsub("_", ".", .)) %>%
summarise_each(funs(min = min(.,na.rm = T),max = max(.,na.rm = T),mean = mean(.,na.rm = T),sd = sd(.,na.rm = T))) %>%
mutate_if(is.numeric, ~round(., 1)) %>%
gather(stat, val,-Innovation,-DelTradMerge) %>%
separate(stat, into = c("Variable", "stat"), sep = "_") %>%
spread(stat,val) %>%
unite(Innovation, Innovation, DelTradMerge, sep = '_')
# A tibble: 54 x 6
Innovation Variable max mean min sd
<chr> <chr> <dbl> <dbl> <dbl> <dbl>
1 Innovators_1 Age 3.5 0.2 0 0.7
2 Innovators_1 Assets 10.7 4.8 1 1.3
3 Innovators_1 Depth 3.5 0.4 0 0.6
4 Innovators_1 Income 7.1 1.3 -6.1 1.3
当我使用此spread(Innovation,-Variable)时，只有min列与Innovation列一起散布，如下所示：
> plot1
# A tibble: 41 x 10
Variable max mean sd Innovators_1 Innovators_2 Innovators_3 NonInnovators_1 NonInnovators_2 NonInnovators_3
<chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Age 2.6 0.2 0.5 NA NA NA 0 NA NA
2 Age 2.8 0.1 0.4 NA 0 NA NA 0 NA
3 Age 3.3 0.2 0.8 NA NA 0 NA NA NA
4 Age 3.3 0.3 0.6 NA NA NA NA NA 0
5 Age 3.5 0.2 0.7 0 NA NA NA NA NA
我也尝试了gather(stat, Innovation, -Variable)，但是它不起作用。
如何同时分散其他统计信息列max mean sd？
这是我的数据：
structure(list(Innovation = c("Innovators_1", "Innovators_1",
"Innovators_1", "Innovators_1", "Innovators_1", "Innovators_1",
"Innovators_1", "Innovators_1", "Innovators_1", "Innovators_2",
"Innovators_2", "Innovators_2", "Innovators_2", "Innovators_2",
"Innovators_2", "Innovators_2", "Innovators_2", "Innovators_2",
"Innovators_3", "Innovators_3", "Innovators_3", "Innovators_3",
"Innovators_3", "Innovators_3", "Innovators_3", "Innovators_3",
"Innovators_3", "NonInnovators_1", "NonInnovators_1", "NonInnovators_1",
"NonInnovators_1", "NonInnovators_1", "NonInnovators_1", "NonInnovators_1",
"NonInnovators_1", "NonInnovators_1", "NonInnovators_2", "NonInnovators_2",
"NonInnovators_2", "NonInnovators_2", "NonInnovators_2", "NonInnovators_2",
"NonInnovators_2", "NonInnovators_2", "NonInnovators_2", "NonInnovators_3",
"NonInnovators_3", "NonInnovators_3", "NonInnovators_3", "NonInnovators_3",
"NonInnovators_3", "NonInnovators_3", "NonInnovators_3", "NonInnovators_3"
), Variable = c("Age", "Assets", "Depth", "Income", "NasdaqIndex",
"Novelty", "PatW", "Sales", "Scope", "Age", "Assets", "Depth",
"Income", "NasdaqIndex", "Novelty", "PatW", "Sales", "Scope",
"Age", "Assets", "Depth", "Income", "NasdaqIndex", "Novelty",
"PatW", "Sales", "Scope", "Age", "Assets", "Depth", "Income",
"NasdaqIndex", "Novelty", "PatW", "Sales", "Scope", "Age", "Assets",
"Depth", "Income", "NasdaqIndex", "Novelty", "PatW", "Sales",
"Scope", "Age", "Assets", "Depth", "Income", "NasdaqIndex", "Novelty",
"PatW", "Sales", "Scope"), max = c(3.5, 10.7, 3.5, 7.1, 8, 0.7,
7.3, 9, 0.7, 2.8, 10.8, 1.5, 7.1, 8, 0.7, 5.6, 10.4, 0.7, 3.3,
10.1, 3, 6.5, 8, 0.7, 6.2, 8.5, 0.7, 2.6, 8.9, 0, 4.4, 8, 0,
0, 8.9, 0, 2.8, 7.9, 0, 3.4, 7.6, 0, 0, 7.7, 0, 3.3, 9.7, 0,
6.3, 8, 0, 0, 9, 0), mean = c(0.2, 4.8, 0.4, 1.3, 6.5, 0.2, 1.5,
3.5, 0.5, 0.1, 4.2, 0.2, 0.9, 6.4, 0.3, 1.4, 3.1, 0.6, 0.2, 5.1,
0.6, 1.7, 6.6, 0.2, 1.9, 3.8, 0.5, 0.2, 3.8, 0, 0.6, 6.4, 0,
0, 2.6, 0, 0.1, 2.8, 0, -0.1, 6.3, 0, 0, 1.8, 0, 0.3, 4, 0, 0.8,
6.5, 0, 0, 2.9, 0), min = c(0, 1, 0, -6.1, 5.7, 0, 0, 0, 0, 0,
0.5, 0, -4.7, 5.7, 0, 0, 0, 0.1, 0, 0.8, 0, -6.4, 5.7, 0, 0,
0.1, 0, 0, 0.6, 0, -6.3, 5.7, 0, 0, 0, 0, 0, 0, 0, -6.3, 5.5,
0, 0, 0, 0, 0, 0.6, 0, -2.8, 5.7, 0, 0, 0, 0), sd = c(0.7, 1.3,
0.6, 1.3, 0.5, 0.2, 1.2, 1.4, 0.2, 0.4, 2.2, 0.3, 1.8, 0.5, 0.2,
1.3, 2.2, 0.2, 0.8, 1.5, 0.6, 1.7, 0.5, 0.2, 1.3, 1.5, 0.2, 0.5,
1.2, 0, 1, 0.5, 0, 0, 1.2, 0, 0.4, 1.4, 0, 1.1, 0.5, 0, 0, 1.4,
0, 0.6, 1.6, 0, 1.3, 0.5, 0, 0, 1.6, 0)), .Names = c("Innovation",
"Variable", "max", "mean", "min", "sd"), class = c("tbl_df",
"tbl", "data.frame"), row.names = c(NA, -54L))
答案 0 :(得分：3)
我找到了此链接https://community.rstudio.com/t/spread-with-multiple-value-columns/5378/2 那里有人用他的功能回答了类似的问题。
myspread <- function(df, key, value) {
# quote key
keyq <- rlang::enquo(key)
# break value vector into quotes
valueq <- rlang::enquo(value)
s <- rlang::quos(!!valueq)
df %>% gather(variable, value, !!!s) %>%
unite(temp, !!keyq, variable) %>%
spread(temp, value)
}
df2 <- df %>% myspread(Innovation, c(max, mean, min, sd))
哪个给出这个结果。
> df2
# A tibble: 27 x 10
DelTradMerge Variable Innovators_max Innovators_mean Innovators_min Innovators_sd NonInnovators_m… NonInnovators_m…
<dbl> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 Age 3.5 0.2 0 0.7 2.6 0.2
2 1 Assets 10.7 4.8 1 1.3 8.9 3.8
3 1 Depth 3.5 0.4 0 0.6 0 0
4 1 Income 7.1 1.3 -6.1 1.3 4.4 0.6
5 1 NasdaqI… 8 6.5 5.7 0.5 8 6.4
6 1 Novelty 0.7 0.2 0 0.2 0 0
7 1 PatW 7.3 1.5 0 1.2 0 0
8 1 Sales 9 3.5 0 1.4 8.9 2.6
9 1 Scope 0.7 0.5 0 0.2 0 0
10 2 Age 2.8 0.1 0 0.4 2.8 0.1
# … with 17 more rows, and 2 more variables: NonInnovators_min <dbl>, NonInnovators_sd <dbl>