在 Mathematica 中将文件内容提取到字符串中的最佳方法是什么？

发布于 2024-09-27 16:17:07 字数 247 浏览 2 评论 0原文

我知道这个问题很常见，但谷歌搜索并没有找到 Mathematica 的明确答案，所以我认为在 StackOverflow 上发布这个问题会很有价值。

我一直在使用 Import 来做这件事，但我突然想到这可能效率非常低，因为 Import 是一个如此重量级的函数。

那么问题来了，你能否在以下方面进行改进：

slurp[filename_] := Import[filename, "Text"]

原文

I know this is asked commonly but googling doesn't turn up a definitive answer for Mathematica so I thought it would be valuable to have this on StackOverflow.

I've been doing this with Import but it occurred to me that that might be horribly inefficient, Import being such a heavyweight function.

So the question is, can you improve on the following:

slurp[filename_] := Import[filename, "Text"]

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野稚 2024-10-04 16:17:07

为了一次导入整个文件，我知道的唯一其他选项是 ReadList。可以将整个文件作为单个字符串返回，如下¹：（

In[1]:= ReadList["ExampleData/source", Record, RecordSeparators -> {}]
Out[1]:= {"f[x] (: function f :)\r\ng[x] (: function g :)\r\n"}

注意：\r 和 \n 实际上在输出中解释，但为了便于阅读，我将它们保留在其中。）是删除任何RecordSeparators。但是，老实说，我认为这不会为您节省任何东西，并且 Import[, "Text"] 更容易编写。说实话，当我的数据格式未包含在 Read 和 中使用的类型说明符中时，我会使用 Read[, String] ReadList，并围绕此操作构建一个自定义函数以加载所有数据。

您可以在阅读文本数据教程中找到此内容。

For importing the entire file at once, the only other option that I am aware of is ReadList. It can be coaxed to returning the entire file as a single string as follows¹:

In[1]:= ReadList["ExampleData/source", Record, RecordSeparators -> {}]
Out[1]:= {"f[x] (: function f :)\r\ng[x] (: function g :)\r\n"}

(Note: \r and \n are actually interpreted in the output, but I left them in for readability.) The key is to remove any RecordSeparators. But, I honestly don't think this saves you anything, and Import[ <file>, "Text"] is easier to write. Truthfully, I use Read[ <file>, String] when I have data in a format that isn't covered by the type specifiers used in Read and ReadList, and build a custom function around this operation to load in all of the data.