我有超过200列的#include "hashTable.h"
#include <algorithm>
hashTable::hashTable()
: tableArray(181667)
{
}
int hashTable::hash(const std::string &nodeData)
{
int multiplier = 1;
int total = 0;
int l = nodeData.length();
for(int i = l - 1; i > -1; --i)
{
int temp = (nodeData[i] - '0') * multiplier;
total += temp;
multiplier *= 10;
}
return total % length;
}
void hashTable::insertNode(const std::string &nodeData)
{
int index = hash(nodeData);
tableArray[index].push_back(nodeData);
}
bool hashTable::removeNode(const string &nodeData)
{
int index = hash(nodeData);
std::list<std::string>::iterator iter = std::find(tableArray[index].begin(), tableArray[index].end(), nodeData);
if (iter != tableArray[index].end())
{
tableArray[index].erase(iter);
return true;
}
return false;
}
bool hashTable::checkForDuplicate(const std::string &nodeData)
{
int index = hash(nodeData);
std::list<std::string>::iterator iter = std::find(tableArray[index].begin(), tableArray[index].end(), nodeData);
return (iter != tableArray[index].end());
}
，并在下面包含了一个子集，其中包含与此问题相关的列：
data.frame我想：
1. 制作样本1-样本10列的所有可能组合，其中每个组合包含来自每个F编号的一个样本，即每个组合包含5个样本，每个样本来自F1，F2，F3， F4，F5。
所以在上面的例子中会有18种组合，例如：
第一个组合是sample1，sample4，sample6，sample7，sample10
第二个组合是sample1，sample4，sample6，sample8，sample10
第三种组合是sample1，sample4，sample6，sample9，sample10
在阅读相关帖子后，我已经使用了>df
Variant Pos ID DB.0.count DB.1.count sample1 sample2 sample3 sample4 sample5 sample6 sample7 sample8 sample9 sample10
variant5 1234567 A 5 5 1/0 1/0 1/0 1/1 1/1 0/0 1/0 0/0 1/0 1/1
. . . . . F1 F1 F1 F2 F2 F3 F4 F4 F4 F5
，unique和duplicated，但却无处可去。
然后，我想将每个唯一组合输出到新的distinct，对样本中的每个变量执行计数，并将结果输出到新列，并执行如下的渔民精确测试并输出到一个新列，如下所示，以下代码应该如下所示:(在这里学习的渔夫代码：Fisher's exact test on values from large dataframe and bypassing errors）
data.frame 2。最后，我想创建一个df.combo.1$pop.0/0.count <- apply(df.combo.1[,6:10], 1, function(u) length(which(grepl("0/0",u))==TRUE) )
df.combo.1$pop.1/0.count <- apply(df.combo.1[,6:10], 1, function(u) length(which(grepl("1/0",u))==TRUE) )
df.combo.1$pop.1/1.count <- apply(df.combo.1[,6:10], 1, function(u) length(which(grepl("1/1",u))==TRUE) )
df.combo.1$pop.0.count <- ( 2*(apply(df.combo.1[,6:10], 1, function(u) length(which(grepl("0/0",u))==TRUE) )) + apply(df.combo.1[,6:10], 1, function(u) length(which(grepl("0/1",u))==TRUE) ) )
df.combo.1$pop.1.count <- ( 2*(apply(df.combo.1[,6:10], 1, function(u) length(which(grepl("1/1",u))==TRUE) )) + apply(df.combo.1[,6:10], 1, function(u) length(which(grepl("0/1",u))==TRUE) ) )
res <- NULL
for (i in 1:nrow(df.combo.1)){
table <- matrix(c(df.combo.1[i, 4], df.combo.1[i, 5], df.combo.1[i, 14], df.combo.1[i, 15]), ncol = 2, byrow = TRUE)
# if any NA occurs in your table save an error in p else run the fisher test
if(any(is.na(table))) p <- "error" else p <- fisher.test(table)$p.value
# save all p values in a vector
res <- c(res,p)
}
df.combo.1$fishers <- res
>df.combo.1
Variant Pos ID DB.0.count DB.1.count sample1 sample4 sample6 sample7 sample10 pop.0/0.count pop.1/0.count pop.1/1.count pop.0.count pop.1.count fishers
variant5 1234567 A 5 5 1/0 1/1 0/0 1/0 1/1 1 2 2 4 6 1.0000
. . . . . F1 F2 F3 F4 F5
，列出每个独特组合的渔民精确p值，如下所示：
data.frame我认为整个练习可能需要某种循环？
答案 0 :(得分：1)
我想我已经掌握了你想要的东西。对于我认为你在第1部分中苦苦挣扎的那一点，我使用了它和expand.grid的组合进行整理。
对于第2部分，一旦数据按每个观察点排列在1行，这是一个相当容易的cbind。
看起来你每次观察使用2行（除非那只是格式化的东西），这使得它非常难（但并非不可能，只需要更多的杂耍），所以我将数据合并到一行。这应该是一个非常简单的转换，只需将每个“第二”行中的相应列附加到每个“第一”行，然后删除每一行。
这可以更有效和整洁地完成，但我认为这是有效的，并且应该相当容易扩展到其他情况。
此致 约什
# provided demo data
# Variant Pos ID DB.0.count DB.1.count sample1 sample2 sample3 sample4 sample5 sample6 sample7 sample8 sample9 sample10
# variant5 1234567 A 5 5 1/0 1/0 1/0 1/1 1/1 0/0 1/0 0/0 1/0 1/1
# . . . . . F1 F1 F1 F2 F2 F3 F4 F4 F4 F5
# create data frame in long format
test.df <- as.data.frame(t(c("variant5",1234567,"A",5,5,"1/0","1/0","1/0","1/1","1/1","0/0","1/0","0/0","1/0","1/1","F1", "F1", "F1", "F2", "F2", "F3", "F4", "F4", "F4", "F5")))
# ensure as character format
test.df[] <- lapply(test.df, as.character)
# get postions of "F" data
F1.var <- which(test.df =="F1")
F2.var <- which(test.df =="F2")
F3.var <- which(test.df =="F3")
F4.var <- which(test.df =="F4")
F5.var <- which(test.df =="F5")
# get all combinations of the 5 F positions
Fcode.combinations <- expand.grid(F1.var,F2.var,F3.var,F4.var,F5.var)
# create results data frame
df.combo.1 <- as.data.frame(matrix(NA,ncol = 21, nrow = nrow(Fcode.combinations)))
# name variables
names(df.combo.1) <- c("Variant","Pos","ID","DB.0.count","DB.1.count",
"F1.sample.pos","F1.result",
"F2.sample.pos","F2.result",
"F3.sample.pos","F3.result",
"F4.sample.pos","F4.result",
"F5.sample.pos","F5.result",
"pop.0_0.count","pop.1_0.count","pop.1_1.count",
"pop.0.count","pop.1.count",
"fishers")
# copy in common data
df.combo.1[,1:5] <- test.df[,1:5]
# setup variables based on combination data
for(i in 1:nrow(Fcode.combinations)){
df.combo.1[i,c(6,8,10,12,14)] <- Fcode.combinations[i,]
# -10 to correct for the position of the results not the 'F type' data
cycle.results <- as.numeric(Fcode.combinations[i,] -10)
df.combo.1[i,c(7,9,11,13,15)] <- test.df[cycle.results]
}
# this is essentially your code with the column reference changed
df.combo.1$pop.0_0.count <- apply(df.combo.1[,c(7,9,11,13,15)], 1, function(u) length(which(grepl("0/0",u))==TRUE) )
df.combo.1$pop.1_0.count <- apply(df.combo.1[,c(7,9,11,13,15)], 1, function(u) length(which(grepl("1/0",u))==TRUE) )
df.combo.1$pop.1_1.count <- apply(df.combo.1[,c(7,9,11,13,15)], 1, function(u) length(which(grepl("1/1",u))==TRUE) )
df.combo.1$pop.0.count <- ( 2*(apply(df.combo.1[,c(7,9,11,13,15)], 1, function(u) length(which(grepl("0/0",u))==TRUE) )) + apply(df.combo.1[,c(7,9,11,13,15)], 1, function(u) length(which(grepl("0/1",u))==TRUE) ) )
df.combo.1$pop.1.count <- ( 2*(apply(df.combo.1[,c(7,9,11,13,15)], 1, function(u) length(which(grepl("1/1",u))==TRUE) )) + apply(df.combo.1[,c(7,9,11,13,15)], 1, function(u) length(which(grepl("0/1",u))==TRUE) ) )
res <- NULL
for (i in 1:nrow(df.combo.1)){
table <- matrix(as.numeric(c(df.combo.1[i, 4], df.combo.1[i, 5], df.combo.1[i, 16], df.combo.1[i, 17])), ncol = 2, byrow = TRUE)
# if any NA occurs in your table save an error in p else run the fisher test
if(any(is.na(table))) p <- "error" else p <- fisher.test(table)$p.value
# save all p values in a vector
res <- c(res,p)
}
df.combo.1$fishers <- res
# create results data
df.combo.1.results <- as.data.frame(cbind(1:nrow(df.combo.1),df.combo.1$fishers))
names(df.combo.1.results) <- c("combo","fishers")