我有两个载体：
c('abc', 'asdf', 'werd', 'ffssd') c('ab', 'd', 'w') 我想对以下for循环进行矢量化：
for(p in 1 : length(patterns)){
count <- count + str_count(texts, p);
}
我使用了以下命令，但两者都不起作用。
> str_count(texts, patterns)
[1] 1 1 1 0
Warning message:
In stri_count_regex(string, pattern, opts_regex = attr(pattern, :
longer object length is not a multiple of shorter object length
> str_count(texts, t(patterns))
[1] 1 1 1 0
Warning message:
In stri_count_regex(string, pattern, opts_regex = attr(pattern, :
longer object length is not a multiple of shorter object length
我想要一个像这样的二维矩阵：
| patterns
------+--------
| 1 0 0
texts | 0 1 0
| 0 1 1
| 0 1 0
答案 0 :(得分：8)
您可以使用<?php
$colors =array("red","green","orange","blue");
$colorLen = count($colors);
?>
<!DOCTYPE html>
<html>
<head>
<link href="stylesheets/reset.css" rel="stylesheet" type="text/css" />
<link href="stylesheets/normalize.css" rel="stylesheet" type="text/css" />
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<title></title>
<style type="text/css">
.myDiv{float:left; width:150px; height:150px; margin:10px;}
.myDiv img{width:100%; height: 100%;}
.myDiv p{text-align: center; margin: 0px; padding: 0px;}
.hidden{display: none;}
.myLink{margin:20px;}
#prev{float: left;}
#next{float: right;}
</style>
<script type="text/javascript">
$(document).ready(function() {
$("#0").removeClass("hidden"); //set the first div to be visible
});
</script>
</head>
<body style="display:inline-block; width:auto;">
<a id="prev" class="myLink" href="#">Prev</a>
<?php //Cycle through the colors and create a div with image and title tag
$i =0;
while ($i != $colorLen) {
echo '<div id="'.$i.'" class="myDiv hidden" style="background-color:'.$colors[$i].';">
<img src="#" alt="'.$colors[$i].'"/><p>'.ucfirst($colors[$i]).'</p>
</div>';
$i++;
}
?>
<a id="next" class="myLink" href="#">Next</a>
<script type="text/javascript">
var numItems =$('.myDiv').length; //Get the number of elements on the page
</script>
</body>
</html>
。我假设您使用outer包中的str_count。
stringr修改
library(stringr)
texts <- c('abc', 'asdf', 'werd', 'ffssd')
patterns <- c('ab', 'd', 'w')
matches <- outer(texts, patterns, str_count)
# set dim names
colnames(matches) <- patterns
rownames(matches) <- texts
matches
ab d w
abc 1 0 0
asdf 0 1 0
werd 0 1 1
ffssd 0 1 0
答案 1 :(得分：3)
使用dplyr和tidyr（以及stringr）：
library(dplyr)
library(tidyr)
library(stringr)
expand.grid(texts, patterns) %>%
mutate_each(funs(as.character(.))) %>%
mutate(matches = stringr::str_count(Var1, Var2)) %>%
spread(Var2, matches)
Var1 ab d w
1 abc 1 0 0
2 asdf 0 1 0
3 ffssd 0 1 0
4 werd 0 1 1