我有一个类似这样的数据框（4行和5列）：
Marker ind1 ind2 ind3 ind4
mark1 CT TT CT TT
mark2 AG AA AG AA
mark3 AC AA AC AA
mark4 CT TT CT TT
我想要做的是将每个列（第一个coloumn除外）拆分为两列。所以输出应该是这样的（4行和9列）：
Marker ind1 ind1 ind2 ind2 ind3 ind3 ind4 ind4
mark1 C T T T C T T T
mark2 A G A A A G A A
mark3 A C A A A C A A
mark4 C T T T C T T T
我知道如何分割一列
do.call(rbind,strsplit(test$JRP4RA6119.039, ""))
给出了这个：
[,1] [,2]
[1,] "C" "T"
[2,] "A" "G"
[3,] "A" "C"
[4,] "C" "T"
我想要的是能够循环这个并为一个数据帧中的所有列创建它。
提前致谢。
答案 0 :(得分：5)
我觉得它有点牵强，但是：
test_split <- data.frame(Marker=test$Marker,
do.call("cbind", lapply(apply(test[, -1], 2, strsplit, ""),
function(x) do.call("rbind", x))),
stringsAsFactors=F)
colnames(test_split)[-1] <- paste(rep(colnames(test)[-1], e=2), 1:2, sep="_")
test_split
# Marker JRP4RA6119.039_1 JRP4RA6119.039_2 JRP4RA6124.029_1 JRP4RA6124.029_2 JRP4RA6133.051_1 JRP4RA6133.051_2 JRP4RA6125.009_1 JRP4RA6125.009_2
#1 s7e4419xxx C T T T C T T T
#2 s7e7001s01 A G A A A G A A
#3 s7e3049xxx A C A A A C A A
#4 s7e4727xxx C T T T C T T T
答案 1 :(得分：5)
您还可以尝试final View parentView = v.findViewById(R.id.touch_delegate_linear_layout);
final View directChildView = v.findViewById(R.id.footer_value_text_view);
ViewTreeObserver vto = parentView.getViewTreeObserver();
vto.addOnGlobalLayoutListener(new ViewTreeObserver.OnGlobalLayoutListener() {
@SuppressLint("NewApi")
@SuppressWarnings("deprecation")
@Override
public void onGlobalLayout() {
final Rect r = new Rect();
final Rect directChild = new Rect();
currencyExchangeLinearLayout.getHitRect(r);
directChildView.getHitRect(directChild);
int bestHeight = Utils.dpToPixel(48);
directChild.top = directChild.bottom - Math.max((bestHeight - r.height()), 0);
parentView.setTouchDelegate(new TouchDelegate(directChild, currencyExchangeLinearLayout));
ViewTreeObserver obs = parentView.getViewTreeObserver();
if (android.os.Build.VERSION.SDK_INT >= android.os.Build.VERSION_CODES.JELLY_BEAN) {
obs.removeOnGlobalLayoutListener(this);
} else {
obs.removeGlobalOnLayoutListener(this);
}
}
});
cSplit_f
splitstackshape或者正如@Ananda Mahto建议的那样，library(splitstackshape)
df1[-1] <- lapply(df1[-1] , function(x)
gsub('(?<=\\w)(?=\\w)', ',', x, perl=TRUE))
cSplit_f(df1, 2:ncol(df1), sep=',')
# Marker ind1_1 ind1_2 ind2_1 ind2_2 ind3_1 ind3_2 ind4_1 ind4_2
#1: mark1 C T T T C T T T
#2: mark2 A G A A A G A A
#3: mark3 A C A A A C A A
#4: mark4 C T T T C T T T
在大型数据集上可能更有效，而且可以直接使用它而无需更改分隔符。
cSplit或使用cSplit(df1, names(df1)[-1], sep="", stripWhite = FALSE)
# Marker ind1_1 ind1_2 ind2_1 ind2_2 ind3_1 ind3_2 ind4_1 ind4_2
#1: mark1 C T T T C T T T
#2: mark2 A G A A A G A A
#3: mark3 A C A A A C A A
#4: mark4 C T T T C T T T
tstrsplit
data.tablelibrary(data.table)#v1.9.5+
setDT(df1)
cbind(Marker=df1$Marker,df1[, unlist(lapply(.SD, function(x)
tstrsplit(x, '')), recursive=FALSE), .SDcols=-1])
# Marker ind11 ind12 ind21 ind22 ind31 ind32 ind41 ind42
#1: mark1 C T T T C T T T
#2: mark2 A G A A A G A A
#3: mark3 A C A A A C A A
#4: mark4 C T T T C T T T
答案 2 :(得分：0)
> b <- as.data.frame(a[, 1])
> b[, 2] <- substr(a[, 2], 1, 1)
> b[, 3] <- substr(a[, 2], 2, 2)
> b[, 4] <- substr(a[, 3], 1, 1)
> b[, 5] <- substr(a[, 3], 2, 2)
> b[, 6] <- substr(a[, 4], 1, 1)
> b[, 7] <- substr(a[, 4], 2, 2)
> b[, 8] <- substr(a[, 5], 1, 1)
> b[, 9] <- substr(a[, 5], 2, 2)
> head(b)
a[, 1] V2 V3 V4 V5 V6 V7 V8 V9
1 mark1 C T T T C T T T
2 mark2 A G A A A G A A
3 mark3 A C A A A C A A
4 mark4 C T T T C T T T
> dim(b)
[1] 4 9
> names(b) <- c("Marker", "ind1", "ind1","ind2", "ind2", "ind3", "ind3", "ind4", "ind4")
> head(b)
Marker ind1 ind1 ind2 ind2 ind3 ind3 ind4
1 mark1 C T T T C T T
2 mark2 A G A A A G A
3 mark3 A C A A A C A
4 mark4 C T T T C T T
ind4
1 T
2 A
3 A
4 T
>
你可以很容易地将它变成一个循环，但是我没有必要使用相对较少的列。
要将其设置为循环，只需将其设置为
即可for(i in 2:ncol(a)){
}