不正确的维数R- rowSums
不正确的维数R- rowSums
提问于 2022-05-18 12:35:50
我对编码非常陌生，所以我基本上是在搜索所有内容，但我无法理解这一条：
我有一个32行和19列的数据框架。我要计算三列中每一行的和。
我是这样写的：
D10 -我的数据帧.
Compliance_score -我要添加的新列
Compliance_1-3 -我想和的列
D10$Compliance_score = rowSums(D10[ ,c("Compliance_1", "Compliance_2", "Compliance_3"), drop = FALSE])我一直收到错误："incorrect number of dimensions".
不知道我做错了什么，或者这个错误信息意味着什么。
有什么想法吗？
**编辑:如果我正确理解了什么是复制的例子(这是我第一次，我希望我做对了-如果没有，请告诉我)
> dput(head(D10))
structure(list(PP = c("003", "014", "047", "013", "053", "048"
), MAAS_1 = c("4.0", "4.0", "3.0", "5.0", "3.0", "4.0"), MAAS_2 =
c("3.0",
"1.0", "6.0", "4.0", "3.0", "3.0"), MAAS_3 = c("4.0", "5.0",
"4.0", "3.0", "4.0", "4.0"), MAAS_4 = c("2.0", "2.0", "6.0",
"2.0", "3.0", "4.0"), MAAS_5 = c("3.0", "3.0", "4.0", "5.0",
"5.0", "5.0"), MAAS_6 = c("3.0", "3.0", "4.0", "3.0", "2.0",
"4.0"), MAAS_7 = c("3.0", "3.0", "4.0", "3.0", "3.0", "5.0"),
MAAS_8 = c("2.0", "4.0", "4.0", "4.0", "4.0", "4.0"), MAAS_9
= c("3.0",
"4.0", "3.0", "2.0", "4.0", "5.0"), MAAS_10 = c("3.0", "4.0",
"4.0", "2.0", "4.0", "4.0"), MAAS_11 = c("2.0", "5.0", "4.0",
"4.0", "1.0", "5.0"), MAAS_12 = c("2.0", "5.0", "6.0", "3.0",
"3.0", "6.0"), MAAS_13 = c("3.0", "3.0", "5.0", "3.0", "3.0",
"2.0"), MAAS_14 = c("3.0", "4.0", "5.0", "4.0", "4.0", "4.0"
), MAAS_15 = c("3.0", "5.0", "6.0", "3.0", "5.0", "5.0"),
Compliance_1 = c("0.0", "0.0", "0.0", "0.0", "1.0", "0.0"
), Compliance_2 = c("1.0", "0.0", "1.0", "0.0", "1.0", "0.0"
), Compliance_3 = c("0.0", "0.0", "0.0", "0.0", "0.0", "0.0"
)), row.names = c(NA, -6L), class = c("tbl_df", "tbl",
"data.frame"
))
>这有意义吗？
回答 2
Stack Overflow用户
发布于 2022-05-18 13:18:54
你可以试试这个：
library(tidyverse)
New_D <- D10 %>%
mutate(Compliance_score = sum(c(Compliance_1, Compliance_2, Compliance_3), na.rm=TRUE))但是，一个可重复的例子将是很好的理解错误。
克莱尔
Stack Overflow用户
发布于 2022-05-19 12:19:31
您的问题是数据存储为字符，因此需要将其转换为类numeric以计算和，即
library(dplyr)
New_D <- df %>%
mutate(across(starts_with("Compliance"), as.numeric)) %>%
mutate(Compliance_score = Compliance_1 + Compliance_2 + Compliance_3)或者在你有NA‘值的时候用“Claire”建议
New_D <- df %>%
mutate(across(starts_with("Compliance"), as.numeric)) %>%
mutate(Compliance_score = sum(c(Compliance_1, Compliance_2, Compliance_3),
na.rm=TRUE))页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/72289301
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