r语言 - 警告消息 在"[<-.factor"("*tmp*"， iseq， value = foo) 中：无效因子水平，尝试将向量添加到行子集时生成的 NA

Question

我正在编写一个函数，该函数尝试一次在多列的 data.frame 的单行中添加值：

require(stringr)
addPointsToKeyRow = function(df, keyRowNum, searchStringForPointColNames, pointsVector){
colsWithMatchingSearchResults = str_match(colnames(df), searchStringForPointColNames)
pointColNums = (which(!is.na(colsWithMatchingSearchResults)))
pointsVectorCleaned = pointsVector[!is.na(pointsVector)]
print(is.vector(pointsVectorCleaned)) #Returns TRUE
print(is.data.frame(pointsVectorCleaned)) #Returns FALSE
print(pointsVectorCleaned)
if(length(pointsVectorCleaned) == length(pointColNums)){
newDf = data.frame(df, stringsAsFactors = FALSE)
newDf[keyRowNum, pointColNums] = as.character(pointsVectorCleaned)
#for(i in 1:length(pointColNums)){
#  newDf[keyRowNum,pointColNums[i]]=as.character(pointsVectorCleaned[i])
#}
print(newDf[keyRowNum,])
}
}

当我将该函数应用于我的数据 (addPointsToKeyRow(finalDf, which(finalDf[,1]=="key"), "points_q", pointVals)( 时，我收到以下警告：

在[<-.factor(*tmp*， iseq， value = "2"( 中： 无效因子水平，NA 生成

我已经在SO和其他网站上查找了错误，建议似乎始终是确保您的data.frame具有stringsAsFactors = FALSE。

我想我的问题可能是当我子集 data.frame (newDf[keyRowNum, pointColNums]时，它不再保持stringsAsFactors = FALSE。

不管这是否是问题所在，我都非常欢迎一些帮助解决这个奇怪的问题。提前非常感谢！

举个例子，假设 df 是：

df = structure(list(first = structure(c(7L, 9L, 5L, 4L, 10L, 2L, 3L, 
6L, 1L, 8L), .Label = c("autumn", "spring", "summer", "winter", 
"july", "betty", "november", "echo", "victor", "tango"), class = "factor"), 
last = structure(c(6L, 2L, 4L, 5L, 1L, 8L, 3L, 9L, 10L, 7L
), .Label = c("brummett1", "do", "drorbaugh", "galeno", "gerber", 
"key", "lyons", "pecsok", "perezfranco", "swatt"), class = "factor"), 
question1 = structure(c(1L, 1L, 1L, 4L, 6L, 2L, 5L, 3L, 5L, 
5L), .Label = c("0", "0.25", "1:02:01", "1:2 50%", "2-Jan", 
"50%"), class = "factor"), points_q1 = structure(c(1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"), 
question2 = structure(c(8L, 10L, 6L, 5L, 2L, 3L, 7L, 1L, 
4L, 9L), .Label = c("        a    |     b; A|    Aa  |  Ab; b|    ab   |  bb; the possibility that the offspring will be heterozygous is about 25%. The same goes for the homozygous recessive it is a 1:1:1:1", 
"1/4 heterozygous for xf1axee and 0 recessive for xf1bxee", 
"16-Mar", "2-Jan", "3:1 25%", "4-Jan", "Male=aabb Female=AAbb Heterozygous is going to be 1/2. Homozygous is going to be 1/4.", 
"possible offspring genotypes (each with probability of 0.25): AABb AaBb AAbb Aabb. Question is asking about probability of Aabb_ which is 0.25.", 
"The square shows Ab Ab_ Bb Bb so 50% or 1/2.  ", "Xa Yb (father) crossed with XA Xb (mother)  = 1/2 "
), class = "factor"), points_q2 = structure(c(1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"), 
question3 = structure(c(4L, 5L, 3L, 5L, 5L, 5L, 7L, 2L, 6L, 
1L), .Label = c("Codominance", "coheritance", "incomplete dominance", 
"Incomplete dominance", "Incomplete dominance ", "Incomplete dominance. ", 
"Independent Assortment"), class = "factor"), points_q3 = structure(c(1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"), 
question4 = structure(c(3L, 4L, 2L, 3L, 6L, 3L, 7L, 1L, 5L, 
4L), .Label = c("", "co-dominance", "Codominance", "Codominance ", 
"Codominance. ", "Codominant ", "Independent Assortment? (Wrong)"
), class = "factor"), points_q4 = structure(c(1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"), 
question5 = structure(c(2L, 10L, 6L, 4L, 5L, 3L, 8L, 1L, 
7L, 9L), .Label = c("      X   |    Y; X|  XX |  XY; x|  Xx  |  xY; the percentage will be 25 % or 1/4 the same applies to the son ", 
"0 for daughter_ because male can only give non-colorblind X chromosome (because he's not colorblind an only has one X chromosome).  0.25 for both son and colorblind.", 
"0.25", "25% for son and 25% for daughter", "25% for the son and 25% for the daughter ", 
"4-Jan", "50%", "Father=XY Mother=X2Y Therefore_ by using the punnet square_ I was able to show/understand that the probability of them having a son AND him being colorblind is 1/4.", 
"To have a son or daughter is 50/50.  To have a colorblind daughter is .25 whereas to have a colorblind son is .75 because it is carried on the X chromosome and the son is much more likely to inherit this because he has less x to work with", 
"XcY (father) XC Xc (mother) Daughter is 1/4 son 1/4"), class = "factor"), 
points_q5 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L), .Label = "", class = "factor"), question6 = structure(c(3L, 
6L, 7L, 8L, 5L, 2L, 10L, 9L, 4L, 1L), .Label = c("Chromatids ", 
"Chromosomes (diploids)", "homologous chromosome pairs", 
"Homologous chromosome pairs are being separated. ", "Homologous chromosomes ", 
"Homologous pairs ", "homologous pairs of chromosomes", "Homologus Chromosomes ", 
"sister chromatids ", "Sister Chromatids?"), class = "factor"), 
points_q6 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L), .Label = "", class = "factor"), question7 = structure(c(6L, 
8L, 5L, 7L, 8L, 2L, 3L, 1L, 9L, 4L), .Label = c("", "Chromatids (haploids)", 
"Daughter Chromosomes?", "One cell to 2", "sister chromatids", 
"Sister chromatids", "Sister Chromatids", "Sister chromatids ", 
"Sister chromatids within daughter cells are separating. "
), class = "factor"), points_q7 = structure(c(1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"), 
question8 = structure(c(1L, 4L, 1L, 2L, 4L, 2L, 3L, 6L, 5L, 
3L), .Label = c("sister chromatids", "Sister chromatids", 
"Sister Chromatids", "Sister chromatids ", "Sister chromatids are held together by the centromeres. In prophase chromosomes become visible. During metaphase chromosomes attach to spindles. During Anaphase the chromosomes are split apart and in telophase the cells start to create cleavage.  ", 
"sisters chromatides"), class = "factor"), points_q8 = structure(c(1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"), 
question9 = structure(c(2L, 4L, 1L, 3L, 4L, 3L, 3L, 2L, 5L, 
3L), .Label = c("prohase ", "prophase", "Prophase", "Prophase ", 
"They condense during prophase before the rest of the phases. "
), class = "factor"), points_q9 = structure(c(1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"), 
question10 = structure(c(1L, 3L, 1L, 2L, 3L, 2L, 2L, 1L, 
4L, 2L), .Label = c("anaphase", "Anaphase", "Anaphase ", 
"During anaphase. "), class = "factor"), points_q10 = structure(c(1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"), 
question11 = structure(c(3L, 4L, 3L, 4L, 4L, 4L, 4L, 3L, 
1L, 2L), .Label = c("During prophase. ", "Telephase ", "telophase", 
"Telophase"), class = "factor"), points_q11 = structure(c(1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"), 
question12 = structure(c(1L, 3L, 1L, 2L, 3L, 2L, 3L, 1L, 
4L, 2L), .Label = c("metaphase", "Metaphase", "Metaphase ", 
"Metaphase. "), class = "factor"), points_q12 = structure(c(1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "", class = "factor"), 
question13 = structure(c(1L, 4L, 1L, 4L, 2L, 4L, 2L, 5L, 
3L, 6L), .Label = c("centromere", "Centromere", "Centromere. ", 
"Centromeres", "centromeres ", "Cleavage"), class = "factor"), 
points_q13 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L), .Label = "", class = "factor")), .Names = c("first", 
"last", "question1", "points_q1", "question2", "points_q2", "question3", 
"points_q3", "question4", "points_q4", "question5", "points_q5", 
"question6", "points_q6", "question7", "points_q7", "question8", 
"points_q8", "question9", "points_q9", "question10", "points_q10", 
"question11", "points_q11", "question12", "points_q12", "question13", 
"points_q13"), row.names = c(NA, -10L), class = "data.frame")

which(finalDf[,1]=="key")为 1。

pointValsc(NA, "2", "2", "2", "2", "2", "2", "2", "1", "1", "1", "1", "1", "1")

为了澄清起见，我希望最终表看起来像这样：

First    Last    question1    points_q1    question2    points_q2    etc.
key    key    0    2    "possible_offspring_genotypes..."    1    etc.

Answer 1

我根据我的理解减少了你的功能，让我知道它是否提供了你想要的东西，或者我是否误解了什么

addPointsToKeyRow = function(df, keyRowNum, searchString, pointsVector) {
#Find columns which has searchString in it
cols <- grepl(searchString, colnames(df))
#Check if the columns with searchString and length of pointsVector is the same
if (sum(cols) == length(pointsVector)) {
#Assign the value
df[keyRowNum,cols] <- pointsVector
}
#Return the updated dataframe
df
}
#Convert all the variables in the column from factor to character
df[] <- lapply(df, as.character)
#define the values to be replaced
pointVals <- c("2", "2", "2", "2", "2", "2", "2", "1", "1", "1", "1","1", "1")
#Call the function
df <- addPointsToKeyRow(df, 1, "points_q", pointsval)
#Check the dataframe
df