如何将功能应用于R中的每一行数据帧？

发布于 2025-01-21 06:21:59 字数 738 浏览 0 评论 0原文

我想将给定函数应用于每一行数据框，并使用其他行值，作为输入参数/参数：

Model <- c("H5", "H5", "H5","H4")
Length <- c(6, 6, 6, 6)
Code <- c("030299", "010121","030448","030324")

df <- data.frame(Model,Length,Code)


Model   Length   Code
HS5       6     030299
HS5       6     010121
HS5       6     030448
HS4       6     030324

我想将以下代码应用于每一行，并将结果作为新列

Library(concordance)

concord(sourcevar = (each row of 'Code' column), origin = as.character(character in 'Model' column) , destination = "HS4", dest.digit = as.numeric(number in 'Length' column), all = F))

文档第6页

原文

I want to apply given function to each row of dataframe and use another values of row, as input parameters/arguments:

Model <- c("H5", "H5", "H5","H4")
Length <- c(6, 6, 6, 6)
Code <- c("030299", "010121","030448","030324")

df <- data.frame(Model,Length,Code)


Model   Length   Code
HS5       6     030299
HS5       6     010121
HS5       6     030448
HS4       6     030324

I want to apply the following code to each row and generate the outcome as a new column

Library(concordance)

concord(sourcevar = (each row of 'Code' column), origin = as.character(character in 'Model' column) , destination = "HS4", dest.digit = as.numeric(number in 'Length' column), all = F))

Documentation Page 6

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夜还是长夜 2025-01-28 06:21:59

您可以在行上使用应用（margin = 1）。

apply(df, MARGIN = 1, function(x) concord(sourcevar = x[3], origin = x[1], destination = "HS4", dest.digit = x[2], all = F))

但是，这是行不通的，因为“ HS4”和“ HS4”之间没有转换字典，因此您只能在不是HS4：的行上使用应用

df$New <- df$Code
df[df$Model != "HS4", ]$New <- apply(df[df$Model != "HS4", ], 1, \(x) concord(sourcevar = x[colnames(df) == "Code"], 
                                               origin = x[colnames(df) == "Model"], destination = "HS4", 
                                               dest.digit = x[colnames(df) == "Length"], all = F))

  Model Length   Code    New
1   HS5      6 030299 030289
2   HS5      6 010121 010121
3   HS5      6 030448 030449
4   HS4      6 030324 030324

You can use apply on the rows (MARGIN = 1).

apply(df, MARGIN = 1, function(x) concord(sourcevar = x[3], origin = x[1], destination = "HS4", dest.digit = x[2], all = F))

However, this does not work because there is no conversion dictionary between "HS4" and "HS4", so you can use apply only on the rows that are not HS4:

df$New <- df$Code
df[df$Model != "HS4", ]$New <- apply(df[df$Model != "HS4", ], 1, \(x) concord(sourcevar = x[colnames(df) == "Code"], 
                                               origin = x[colnames(df) == "Model"], destination = "HS4", 
                                               dest.digit = x[colnames(df) == "Length"], all = F))

  Model Length   Code    New
1   HS5      6 030299 030289
2   HS5      6 010121 010121
3   HS5      6 030448 030449
4   HS4      6 030324 030324

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