我从人类蛋白质图谱下载了一个数据集,其中包含12,004种蛋白质的亚细胞定位注释。这个文件,我有子集,只包括“基因名称”,然后4列是如何可靠的位置(基于免疫荧光染色的细胞)。这些是"Validated">"Supported">"Approved">"Uncertain".
我想出了一个评分系统,我想应用于LC光谱计数数据集,我有1)权衡注释的质量,2)惩罚在拟议评分系统的图像中发现蛋白质的位置。
TLDR是指我需要计算以下数据集的每一列中有多少项,并获得该信息的数据。
df <- read.csv("proteinAtlas.csv")
dput(df)
structure(list(Gene_symbol = structure(1:49, .Label = c("AAAS",
"AAMP", "AAR2", "AARD", "AARS", "AARS2", "AARSD1", "ABCA13",
"ABCB6", "ABCB7", "ABCB8", "ABCC1", "ABCC4", "ABCD3", "ABCE1",
"ABCF1", "ABCF2", "ABCF3", "ABHD10", "ABHD14B", "ABHD6", "ABI1",
"ABI2", "ABL2", "ACAA1", "ACAA2", "ACACA", "ACAD9", "ACADM",
"ACADS", "ACADVL", "ACAP1", "ACAP2", "ACAT1", "ACAT2", "ACBD3",
"ACBD5", "ACIN1", "ACLY", "ACO2", "ACOT1", "ACOT13", "ACOT2",
"ACOT7", "ACOT8", "ACOT9", "ACOX1", "ACP1", "ACP5"), class = "factor"),
Validated = structure(c(1L, 2L, 1L, 1L, 2L, 4L, 1L, 1L, 3L,
1L, 1L, 1L, 1L, 5L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
5L, 1L, 1L, 4L, 4L, 1L, 1L, 1L, 1L, 4L, 1L, 1L, 5L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 6L, 1L, 1L), .Label = c("", "Cytosol",
"Golgi apparatus", "Mitochondria", "Peroxisomes", "Vesicles"
), class = "factor"), Supported = structure(c(1L, 9L, 1L,
1L, 1L, 1L, 1L, 1L, 5L, 10L, 10L, 12L, 1L, 1L, 1L, 1L, 4L,
1L, 1L, 6L, 1L, 3L, 1L, 11L, 1L, 10L, 2L, 1L, 1L, 10L, 10L,
1L, 1L, 1L, 4L, 8L, 1L, 11L, 7L, 10L, 1L, 1L, 1L, 4L, 13L,
1L, 1L, 1L, 1L), .Label = c("", "Actin filaments;Cytosol",
"Cell Junctions;Plasma membrane", "Cytosol", "Cytosol;Mitochondria;Nucleoplasm;Plasma membrane",
"Cytosol;Nucleoli;Nucleus", "Cytosol;Nucleoplasm;Plasma membrane",
"Golgi apparatus", "Microtubules", "Mitochondria", "Nucleoplasm",
"Plasma membrane", "Vesicles"), class = "factor"), Approved = structure(c(3L,
1L, 5L, 12L, 1L, 1L, 6L, 4L, 1L, 1L, 17L, 1L, 8L, 1L, 1L,
1L, 1L, 7L, 13L, 1L, 16L, 1L, 15L, 1L, 1L, 1L, 14L, 1L, 1L,
15L, 17L, 18L, 11L, 1L, 17L, 1L, 1L, 1L, 1L, 1L, 13L, 2L,
13L, 15L, 13L, 9L, 17L, 10L, 5L), .Label = c("", "Cell Junctions",
"Centrosome;Cytosol;Nuclear membrane", "Centrosome;Cytosol;Vesicles",
"Cytosol", "Cytosol;Nuclear membrane", "Cytosol;Nucleoli",
"Cytosol;Nucleoli;Plasma membrane", "Cytosol;Nucleoplasm;Plasma membrane",
"Cytosol;Nucleus", "Endosomes", "Lipid droplets", "Mitochondria",
"Nucleoli fibrillar center", "Nucleoplasm", "Nucleoplasm;Vesicles",
"Nucleus", "Vesicles"), class = "factor"), Uncertain = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L,
1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L), .Label = c("", "Cytosol;Plasma membrane", "Nucleoli"
), class = "factor")), .Names = c("Gene_symbol", "Validated",
"Supported", "Approved", "Uncertain"), class = "data.frame", row.names = c(NA,
-49L))因此理想的输出应该类似于这个图,或者,如果您愿意的话,dput():
structure(list(Gene_symbol = structure(1:29, .Label = c("AAAS",
"AAMP", "AAR2", "AARD", "AARS", "AARS2", "AARSD1", "ABCA13",
"ABCB6", "ABCB7", "ABCB8", "ABCC1", "ABCC4", "ABCD3", "ABCE1",
"ABCF1", "ABCF2", "ABCF3", "ABHD10", "ABHD14B", "ABHD6", "ABI1",
"ABI2", "ABL2", "ACAA1", "ACAA2", "ACACA", "ACAD9", "ACADM"), class = "factor"),
Validated = c(NA, 1L, NA, NA, 1L, 1L, NA, NA, 1L, NA, NA,
NA, NA, 1L, 1L, 1L, NA, NA, NA, NA, NA, NA, NA, NA, 1L, NA,
NA, 1L, 1L), Supported = c(NA, 1L, NA, NA, NA, NA, NA, NA,
4L, 1L, 1L, 1L, NA, NA, NA, NA, 1L, NA, NA, 3L, NA, 2L, NA,
1L, NA, 1L, 2L, NA, NA), Approved = c(3L, NA, 1L, 1L, NA,
NA, 2L, 3L, NA, NA, 1L, NA, 3L, NA, NA, NA, NA, 2L, 1L, NA,
2L, NA, 1L, NA, NA, NA, 1L, NA, NA), Uncertain = c(NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), .Names = c("Gene_symbol",
"Validated", "Supported", "Approved", "Uncertain"), class = "data.frame", row.names = c(NA,
-29L))在每一列中,大部分是由";“分隔的字符串;然而,在某些情况下,它们是像”原子核纤维中心“或”脂滴“这样的术语,它们由空格隔开,应该算作一个单词/术语。
我发现了计算R中字符串中的单词数的例子,其中:
d <- "foo,bar,fun"
length(strsplit(d,",")[[1]]
class(d)但这只适用于“字符”类,而不适用于"data.frame“。
有人能建议如何在R中这样做吗?非常感谢!
发布于 2017-11-19 17:37:18
我们可以使用str_count。循环除第一个列(lapply(df[-1], ..)以外的列,获取;添加1的计数,检查是否存在空字符串,并使用NA替换这些元素。
library(stringr)
df[-1] <- lapply(df[-1], function(x) (str_count(x, ";") + 1) * NA^(as.character(x) == ""))发布于 2017-11-19 17:48:25
一种使用base的解决方案
result_df <- data.frame(t(apply(df,1,function(x){
c(x[1],sapply(strsplit(as.character(x[-1]),";"),length))
})), stringsAsFactors = F)
names(result_df) <- c("Gene_symbol", "Validated", "Supported", "Approved", "Uncertain")https://stackoverflow.com/questions/47379863
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