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在 Mathematica 中重新排序列表元素的更好方法

发布于 2024-12-23 00:43:26 字数 187 浏览 1 评论 0原文

我想根据每个元素的第一个组件按 lst = {"3", "1", “2”，“9”}。也就是说，我希望输出为 {{{"3", 2}}, {{"1", 2}}, {{"2", 2}}, {{"9", 2 }}}。

目前我有一种丑陋的解决方案，但想知道是否有人可以帮助提供一个简洁的解决方案。多谢。

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赏烟花じ飞满天 2024-12-30 00:43:26

一种方法可能是

data = Tally[Characters["2723198931"]];
lst = {"3", "1", "2", "9"};

(*algorithm*)

pos =  Position[data[[All, 1]], #] & /@ lst;
Extract[data, pos]

输出

{{{"3", 2}}, {{"1", 2}}, {{"2", 2}}, {{"9", 2}}}

上面的更新 2:37 am

屏幕截图，我在 Windows 7 上使用 v 8.04

在此输入图像描述

one way may be

data = Tally[Characters["2723198931"]];
lst = {"3", "1", "2", "9"};

(*algorithm*)

pos =  Position[data[[All, 1]], #] & /@ lst;
Extract[data, pos]

output

{{{"3", 2}}, {{"1", 2}}, {{"2", 2}}, {{"9", 2}}}

Update 2:37 am

screen shot of the above, I am using v 8.04 on windows 7

enter image description here

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没有伤那来痛 2024-12-30 00:43:26

这应该是相当快的，但我没有测试过。在长列表中，您可以使用 Dispatch[rls] 获得更好的性能。

key = {"3", "1", "2", "9"};
tal = Tally[Characters["2723198931"]];

rls = #[[1]] -> # & /@ tal;
key /. rls

{{"3", 2}, {"1", 2}, {"2", 2}, {"9", 2}}

或者，您可以使用它，这对于长 Tally 列表来说速度要快一些：

rls = Thread[tal[[All, 1]] -> tal];

This should be quite fast, but I didn't test it. On long lists you may get better performance using Dispatch[rls].

key = {"3", "1", "2", "9"};
tal = Tally[Characters["2723198931"]];

rls = #[[1]] -> # & /@ tal;
key /. rls

{{"3", 2}, {"1", 2}, {"2", 2}, {"9", 2}}

Alternatively you could use this, which is little faster for long Tally lists:

rls = Thread[tal[[All, 1]] -> tal];

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一片旧的回忆 2024-12-30 00:43:26

或者，您可以执行类似的操作，

With[{tal = Tally[Characters["2723198931"]]},
  Flatten@Pick[tal, tal[[All, 1]], #] & /@ key]

您是否总是计算字符串中的字符？如果是这样，直接计算字符数可能比先将其转换为字符列表更有效。例如考虑

str = StringJoin[RandomChoice[CharacterRange["0", "9"], 1000]];
key = {"3", "1", "2", "9"};

({#, StringCount[str, #]} & /@ key) // Timing

(* output: {0.000121, {{"3", 98}, {"1", 112}, {"2", 99}, {"9", 107}}} *)

With[{tal = Tally[Characters[str]]},
  Flatten@Pick[tal, tal[[All, 1]], #] & /@ key] // Timing 

(* output: {0.000567, {{"3", 98}, {"1", 112}, {"2", 99}, {"9", 107}}} *)

Alternatively, you could do something like

With[{tal = Tally[Characters["2723198931"]]},
  Flatten@Pick[tal, tal[[All, 1]], #] & /@ key]

Are you always counting characters in a string? If so it might be more efficient to count the characters directly instead of transforming it to a list of characters first. Consider for example

str = StringJoin[RandomChoice[CharacterRange["0", "9"], 1000]];
key = {"3", "1", "2", "9"};

({#, StringCount[str, #]} & /@ key) // Timing

(* output: {0.000121, {{"3", 98}, {"1", 112}, {"2", 99}, {"9", 107}}} *)

With[{tal = Tally[Characters[str]]},
  Flatten@Pick[tal, tal[[All, 1]], #] & /@ key] // Timing 

(* output: {0.000567, {{"3", 98}, {"1", 112}, {"2", 99}, {"9", 107}}} *)

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~没有更多了~