JS過濾數組數組

Question

我正在嘗試過濾數組數組。

我已經搜索過SO的答案，但是遇到的所有問題都與我的不匹配（對象數組或未嵌套或格式不相同的簡單數組，等等...）

將值存儲在數組中后，它們將如下所示：

[[Paris, One ONE, Boss, Wed Mar 01 00:00:00 GMT+01:00 2017, ], [Paris, Two TWO, Temp, Sat Jul 01 00:00:00 GMT+02:00 2017, ], [Paris, Three THREE, Employee, Sat Sep 01 00:00:00 GMT+02:00 2018, ], [Paris, Four FOUR, Intern, Thu Nov 01 00:00:00 GMT+01:00 2018, ], [Paris, Five FIVE, N.A., Sat Dec 01 00:00:00 GMT+01:00 2018, ], [Paris, Six SIX, Director, Tue Jan 01 00:00:00 GMT+01:00 2019, ], [Paris, Seven SEVEN, Director, Fri Jan 01 00:00:00 GMT+01:00 2016, Sun Jul 01 00:00:00 GMT+02:00 2018], [Paris, Eight EIGHT, Director, Fri Jan 01 00:00:00 GMT+01:00 2016, Sun Oct 01 00:00:00 GMT+02:00 2017], [Paris, Nine NINE, N.A., Thu Nov 01 00:00:00 GMT+01:00 2018, Sat Dec 01 00:00:00 GMT+01:00 2018], [London, Ten TEN, Employee, Fri Jan 01 00:00:00 GMT+01:00 2016, Mon Oct 01 00:00:00 GMT+02:00 2018], [London, Eleven ELEVEN, Intern, Mon Feb 01 00:00:00 GMT+01:00 2016, Mon Jan 01 00:00:00 GMT+01:00 2018], [London, Twelve TWELVE, Employee, Sun May 01 00:00:00 GMT+02:00 2016, Sun Oct 01 00:00:00 GMT+02:00 2017]]

我希望能夠過濾此數組數組，並使數據僅鏈接到一個社區，例如巴黎。

我該怎么做？

非常感謝你的幫助

Answer 1

您可以使用Array.filter這樣的東西：

 const data = [ ['Paris', 'One ONE', 'Boss', 'Wed Mar 01 00:00:00 GMT+01:00 2017' ], ['Paris', 'Two TWO', 'Temp', 'Sat Jul 01 00:00:00 GMT+02:00 2017' ], ['London', 'Three THREE', 'Employee, Sat Sep 01 00:00:00 GMT+02:00 2018' ]]; const result = data.filter(function(item) { return item[0]==='Paris'}); // const result = data.filter(item => item[0]==='Paris'); // ES6 console.log(result); 

Answer 2

假設數組中的所有元素都是Strings ，則只需檢查數組中的每個數組都包含元素Paris 。

 const yourArray = [ ['Paris', 'one One', 'whatever else'], ['Paris', 'one One', 'whatever else'], ['Paris', 'one One', 'whatever else'], ['Paris', 'one One', 'whatever else'], ['London', 'one One', 'whatever else'], ] const onlyParis = yourArray.filter(function(array) { return array.includes('Paris') }) console.log(onlyParis) 

Answer 3

我已經將您的數組轉換為JavaScript數組。 過濾器將像這樣完成：

 var myArray = [['Paris', 'One ONE', 'Boss', 'Wed Mar 01 00:00:00 GMT+01:00 2017'],['Paris', 'Two TWO', 'Temp', 'Sat Jul 01 00:00:00 GMT+02:00 2017'],['Paris', 'Three THREE', 'Employee', 'Sat Sep 01 00:00:00 GMT+02:00 2018'],['Paris', 'Four FOUR', 'Intern', 'Thu Nov 01 00:00:00 GMT+01:00 2018'],['Paris', 'Five FIVE', 'NA', 'Sat Dec 01 00:00:00 GMT+01:00 2018'],['Paris', 'Six SIX', 'Director', 'Tue Jan 01 00:00:00 GMT+01:00 2019'],['Paris', 'Seven SEVEN', 'Director', 'Fri Jan 01 00:00:00 GMT+01:00 2016', 'Sun Jul 01 00:00:00 GMT+02:00 2018'],['Paris', 'Eight EIGHT', 'Director', 'Fri Jan 01 00:00:00 GMT+01:00 2016', 'Sun Oct 01 00:00:00 GMT+02:00 2017'],['Paris', 'Nine NINE', 'NA', 'Thu Nov 01 00:00:00 GMT+01:00 2018', 'Sat Dec 01 00:00:00 GMT+01:00 2018'],['London', 'Ten TEN', 'Employee', 'Fri Jan 01 00:00:00 GMT+01:00 2016', 'Mon Oct 01 00:00:00 GMT+02:00 2018'],['London', 'Eleven ELEVEN', 'Intern', 'Mon Feb 01 00:00:00 GMT+01:00 2016', 'Mon Jan 01 00:00:00 GMT+01:00 2018'],['London', 'Twelve TWELVE', 'Employee', 'Sun May 01 00:00:00 GMT+02:00 2016', 'Sun Oct 01 00:00:00 GMT+02:00 2017']] var filteredArray = myArray.filter(function (item){ return item[0] == 'Paris' }) console.log(filteredArray) 

Answer 4

此功能過濾列表的lst元素。 它比較每個列表的第一個元素（在您的示例中為城市名稱），並且僅當它等於“巴黎”時才返回true。

這隱含了一個假設，即innerLst的結構不會改變。 如果將城市名稱移到其他索引，則larz的解決方案會解決這個問題。

lst.filter(innerLst => innerLst[0] === 'Paris')

Answer 5

 const country = [ ["Paris", "One ONE, Boss, Wed Mar 01 00: 00: 00 GMT + 01: 00 2017", ], ["Paris", "Two TWO, Temp, Sat Jul 01 00: 00: 00 GMT + 02: 00 2017", ], ["Paris", "Three THREE, Employee, Sat Sep 01 00: 00: 00 GMT + 02: 00 2018", ], ["Paris", "Four FOUR, Intern, Thu Nov 01 00: 00: 00 GMT + 01: 00 2018", ], ["Paris", "Five FIVE, NA, Sat Dec 01 00: 00: 00 GMT + 01: 00 2018", ], ["Paris, Seven SEVEN, Director, Fri Jan 01 00: 00: 00 GMT + 01: 00 2016, Sun Jul 01 00: 00: 00 GMT + 02: 00 2018"], ["London", "Twelve TWELVE, Employee, Sun May 01 00: 00: 00 GMT + 02: 00 2016", "Sun Oct 01 00: 00: 00 GMT + 02: 00 2017"] ] const res = country.filter(([ville, b, c]) => ville === "Paris") console.log(res)