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App Inventor 2 图片云端保存及访问的开发思路 · App Inventor 2 中文网

... 教育 入门必读 中文教程 IoT专题 AI2拓展 ChatGPT接入 Aia Store 开通VIP 搜索 App Inventor 2 图片云端保存及访问的...
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App Inventor 2 实现蓝牙未开启时弹窗提醒用户开启蓝牙 · App Inventor 2 中文网

... 教育 入门必读 中文教程 IoT专题 AI2拓展 ChatGPT接入 Aia Store 开通VIP 搜索 App Inventor 2 实现蓝牙未开启时弹窗...
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MIT官方已升级至2.73版本,中文网待测试并升级相关特性 - App Inventor 2 ...

...版本重点更新 Android Companion 应用及其底层框架。新版本 2.73 现已可从 Google Play 商店和 ai2.appinventor.mit.edu(版本 2.73u)直接下载。 更改: 为 ListView 组件实现“提示”属性,并全面提高其性能。 在 ChatBot 组件中添加下拉菜...
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App Inventor 2 ECharts 拓展:基于 ECharts 强大的个性化数据图表展示 · ...

... 教育 入门必读 中文教程 IoT专题 AI2拓展 ChatGPT接入 Aia Store 开通VIP 搜索 App Inventor 2 ECharts 拓展:基于 ECharts ...
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【免费开放】App Inventor 2 LLMAI2Ext 自研拓展:接入DeepSeek、Kimi、通...

最新aix拓展永久下载地址:https://www.fun123.cn/reference/extensions/LLMAI2Ext.html 中文网开发国内大模型拓展的初衷 App Inventor 2 原生的ChatGPT组件由于是国外的,使用起来不太便捷,且各种限制。如今我们又身处AI浪潮之中,包括很多...
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Best way to merge two maps and sum the values of same key?

...> import Scalaz._ import Scalaz._ scala> val map1 = Map(1 -> 9 , 2 -> 20) map1: scala.collection.immutable.Map[Int,Int] = Map(1 -> 9, 2 -> 20) scala> val map2 = Map(1 -> 100, 3 -> 300) map2: scala.collection.immutable.Map[Int,Int] = Map(1 -> 100, 3 -> 300) scala&g...
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Return first N key:value pairs from dict

... | edited Aug 21 '19 at 12:43 ofir_aghai 1,89811 gold badge2727 silver badges3030 bronze badges ...
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Javascript equivalent of Python's zip function

... 2016 update: Here's a snazzier Ecmascript 6 version: zip= rows=>rows[0].map((_,c)=>rows.map(row=>row[c])) Illustration equiv. to Python{zip(*args)}: > zip([['row0col0', 'row0col1', 'row0col2'], ['row1c...
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How to calculate a Mod b in Casio fx-991ES calculator

... Now do your calculation (in comp mode), like 50 / 3 and you will see 16 2/3, thus, mod is 2. Or try 54 / 7 which is 7 5/7 (mod is 5). If you don't see any fraction then the mod is 0 like 50 / 5 = 10 (mod is 0). The remainder fraction is shown in reduced form, so 60 / 8 will result in 7 1/2. Rema...
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combinations between two lists?

...e looking for - see the other answers. Suppose len(list1) >= len(list2). Then what you appear to want is to take all permutations of length len(list2) from list1 and match them with items from list2. In python: import itertools list1=['a','b','c'] list2=[1,2] [list(zip(x,list2)) for x in ite...