blocks|key|4297042|text|首先我将生成随机数据来测试这一点|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4297043|#+generate+random+data+
test_data+<-+data.frame(x+=+1:100,+y+=+rnorm(100))
#+add+random+NAs
test_data$y[sample(1:100,+50)]+<-+NA|code-block|syntax|javascript|4297044|现在试试这个：|4297045|#+locate+non+NAs+in+the+wanted+column
not_na+<-+which(!is.na(test_data$y))

#+define+the+function+replace_NAs_custom

replace_NAs_custom+<-+function(i,+col){
+++++++++if(is.na(col[i])){
+++++++++++col[i]+<-+col[max(not_na[not_na+<+i]+)]
+++++++++}
+++++++++return(col[i]+)
+++++++}

test_data$y_2+<-+unlist(lapply(1:nrow(test_data),+replace_NAs_custom,+test_data$y))|4297046|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

First I will generate random data to test this

<pre><code># generate random data 
test_data &lt;- data.frame(x = 1:100, y = rnorm(100))
# add random NAs
test_data$y[sample(1:100, 50)] &lt;- NA
</code></pre>

Now try this:

<pre><code># locate non NAs in the wanted column
not_na &lt;- which(!is.na(test_data$y))

# define the function replace_NAs_custom

replace_NAs_custom &lt;- function(i, col){
 if(is.na(col[i])){
 col[i] &lt;- col[max(not_na[not_na &lt; i] )]
 }
 return(col[i] )
 }

test_data$y_2 &lt;- unlist(lapply(1:nrow(test_data), replace_NAs_custom, test_data$y))
</code></pre>

blocks|key|4297103|text|看起来你的代码在每次循环时都会进行大量的计算和内存分配。为了减少时间，我们希望减少循环在每次迭代中所做的工作量。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4297104|我不是百分之百清楚你的问题，但我想我已经掌握了要点。听起来您只想获取最后一个非NA值并将其复制到包含NA值的行中。我们可以使用一对或索引来做到这一点。|4297105|在下面的方法中，在我进入循环之前，所有的内存都已经预先分配好了。惟一的内存操作是将一个值(NA)替换为另一个值。除了该操作之外，还需要检查该值是否为NA，并对索引执行加法操作。为了更快地解决这个问题，您需要使用c优化的向量函数(可能来自包/库)。|4297106|要使用前一个值填充NA：|4297107|#+Fill+with+previous+non-NA+value
VERINDEX_VEC+<-+c(NA,"A1","A2",NA,NA,"A3",NA)
VERINDEX_VEC
#+[1]+NA+++"A1"+"A2"+NA+++NA+++"A3"+NA++

non_na_positions+<-+which(!is.na(VERINDEX_VEC))
#+If+the+first+value+is+NA+we+need+to+fill+with+NA+until+we+hit+a+known+value...
if(is.na(VERINDEX_VEC[1])){
++non_na_positions+<-+c(NA,non_na_positions)
}

index+=+1

for(i+in+1:length(VERINDEX_VEC)){
++if(is.na(VERINDEX_VEC[i]))+{
++++VERINDEX_VEC[i]+<-+VERINDEX_VEC[non_na_positions[index]]
++}+else+{
++++index+<-+index+%2B+1
++}
}

VERINDEX_VEC
#+[1]+NA+++"A1"+"A2"+"A2"+"A2"+"A3"+"A3"|code-block|syntax|javascript|4297108|使用下一个值填充NA：|4297109|#+Fill+with+next+non-NA+Value
VERINDEX_VEC+<-+c(NA,"A1","A2",NA,NA,"A3",NA)
VERINDEX_VEC
#+[1]+NA+++"A1"+"A2"+NA+++NA+++"A3"+NA++

non_na_positions+<-+which(!is.na(VERINDEX_VEC))
#+Never+need+the+first+position+of+the+vector+if+we+are+looking-ahead...
index+<-+ifelse(non_na_positions[1]==1,2,1)

for(i+in+1:length(VERINDEX_VEC)){
++if(is.na(VERINDEX_VEC[i]))+{
++++VERINDEX_VEC[i]+<-+VERINDEX_VEC[non_na_positions[index]]
++}+else+{
++++index+<-+index+%2B+1
++}
}

VERINDEX_VEC
#+[1]+"A1"+"A1"+"A2"+"A3"+"A3"+"A3"+NA++|4297110|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|U|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|V|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

It looks like your code is doing a lot of calculating and memory allocation every time it loops. To decrease time we want to decrease how much work the loop does each iteration. 

I'm not 100% clear on your problem, but I think I've got the gist of it. It sounds like you just want to take the last non-NA value and copy it into the row with the NA value. We can use a pair or indexes to do this. 

In the following method all of the memory is already pre-allocated before I enter the loop. The only memory action is to replace a value (NA) with another value. Other then that operation there is a check to see if the value is NA and there is an addition operation on the index. In order to get significantly faster on this problem you would need to use c-optimized vector functions (probably from a package/library). 

To use the previous value to fill NA:

<pre><code># Fill with previous non-NA value
VERINDEX_VEC &lt;- c(NA,"A1","A2",NA,NA,"A3",NA)
VERINDEX_VEC
# [1] NA "A1" "A2" NA NA "A3" NA 

non_na_positions &lt;- which(!is.na(VERINDEX_VEC))
# If the first value is NA we need to fill with NA until we hit a known value...
if(is.na(VERINDEX_VEC[1])){
 non_na_positions &lt;- c(NA,non_na_positions)
}

index = 1

for(i in 1:length(VERINDEX_VEC)){
 if(is.na(VERINDEX_VEC[i])) {
 VERINDEX_VEC[i] &lt;- VERINDEX_VEC[non_na_positions[index]]
 } else {
 index &lt;- index + 1
 }
}

VERINDEX_VEC
# [1] NA "A1" "A2" "A2" "A2" "A3" "A3"
</code></pre>

To use the next value to fill NA:

<pre><code># Fill with next non-NA Value
VERINDEX_VEC &lt;- c(NA,"A1","A2",NA,NA,"A3",NA)
VERINDEX_VEC
# [1] NA "A1" "A2" NA NA "A3" NA 

non_na_positions &lt;- which(!is.na(VERINDEX_VEC))
# Never need the first position of the vector if we are looking-ahead...
index &lt;- ifelse(non_na_positions[1]==1,2,1)

for(i in 1:length(VERINDEX_VEC)){
 if(is.na(VERINDEX_VEC[i])) {
 VERINDEX_VEC[i] &lt;- VERINDEX_VEC[non_na_positions[index]]
 } else {
 index &lt;- index + 1
 }
}

VERINDEX_VEC
# [1] "A1" "A1" "A2" "A3" "A3" "A3" NA 
</code></pre>

I am trying to fill the NA values with the non-NA previous available in a set of data of 3 million rows. At the moment I am able to do it but it takes about 3hrs.

Constrains - I cannot use any library, it has to be done with R basic

Data - My data looks as following (Extract)

<a href="https://i.stack.imgur.com/wr9LN.png" rel="nofollow noreferrer">Data Extract As Example</a>

At the moment I have been using the following code

<pre><code>CHARDIF &lt;- diff(VERINDEX_VEC)

k = 1
for (j in VERINDEX_VEC){
 #when value is in vector calculate difference to next value and copy VER. 
Special cases for First and Last value
 ifelse(j == 1, ALL_POS$C01[j:CHARDIF[k]] &lt;- ALL_POS$C01[j],
 ifelse(j == max(VERINDEX_VEC), ALL_POS$C01[j:max(as.numeric
(row.names(ALL_POS)))] &lt;- ALL_POS$C01[j],ALL_POS$C01[j:(j+CHARDIF[k]-1)] &lt;- 
ALL_POS$C01[j]))
 k = k + 1
}
</code></pre>

As you can see I have a vector with the non-NA positions, and then I calculate the difference between the positions, and that helps me to select the range I want to paste as I know when the next non-NA value is happening.

Does anyone have a better solution? a faster one in particular

Fill NAs with Previous non-NA available value R

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在尝试用一组300万行数据中以前可用的非NA值填充NA值。目前我可以做到这一点，但这需要大约3小时。Constrains -我不能使用任何库，它必须用R basic完成数据-我的数据如下所示(摘录)目前，我一直在使用以下代码CHARDIF <- diff(VERINDEX_VEC)k = 1for (j in VE...

问用以前的非NA可用值R填充NAs
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问用以前的非NA可用值R填充NAs
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