Question

删除列表中所有列表的最pythonic方法是什么？

例如，如果有一个像[1,2,['randompie'],3,[],4,5]这样的列表，我该如何制作这样的[1,2,3,4,5]

这是我尝试过的：

[elem for elem in [1,2,['randompie'],3,[],4,5] if type(elem)!='list']

Answer 1

我会使用列表理解：

import 'dart:async';
import 'package:flutter/material.dart';

void main() {
  runApp(new MaterialApp(
    home: new Scaffold(
      appBar: new AppBar(title: new Text('Example App')),
      body: new textList(),
    ),
  ));
}

class textList extends StatefulWidget {

  @override
  State<StatefulWidget> createState() =>
      new _textListState();
}

class _textListState extends State<textList>
    with TickerProviderStateMixin {

  List<Widget> items = new List();
  Widget lorem = new textClass("Lorem");
  Timer timer;

  @override
  void initState() {
    super.initState();

    items.add(new textClass("test"));
    items.add(new textClass("test"));

    timer = new Timer.periodic(new Duration(seconds: 5), (Timer timer) {
      setState(() {
        items.removeAt(0);
        items.add(lorem);
      });
    });
  }

  @override
  void dispose() {
    super.dispose();
    timer.cancel();
  }

  @override
  Widget build(BuildContext context) {
    Iterable<Widget> content = ListTile.divideTiles(
        context: context, tiles: items).toList();

    return new Column(
      children: content,
    );
  }
}

class textClass extends StatefulWidget {
  textClass(this.word);

  final String word;

  @override
  State<StatefulWidget> createState() =>
      new _textClass();
}

class _textClass extends State<textClass>
    with TickerProviderStateMixin {
  _textClass();

  String word;
  Timer timer;

  @override
  void didUpdateWidget(textClass oldWidget) {
    if (oldWidget.word != widget.word) {
      word = widget.word;
    }
    super.didUpdateWidget(oldWidget);
  }

  @override
  void initState() {
    super.initState();
    word = widget.word;

    timer = new Timer.periodic(new Duration(seconds: 2), (Timer timer) {
      setState(() {
        word += "t";
      });
    });
  }

  @override
  void dispose() {
    super.dispose();
    timer.cancel();
  }


  @override
  Widget build(BuildContext context) {
    return new Text(word);
  }
}

Answer 2

您可以使用列表推导来轻松过滤列表中的所有非列表元素：

>>> l = [1,2,['randompie'],3,[],4,5]
>>> [el for el in l if not isinstance(el, list)]
[1, 2, 3, 4, 5]

注意我使用了isinstance而不是type。这有两个原因。前一个函数是首选，因为it takes parent classes into account。因为isinstance允许您轻松扩展列表理解以过滤掉其他类型，例如tuple s或dict s：

>>> l = [1, 2, ['randompie'], 3, [], 4, 5, (1,)]
>>> [el for el in l if not isinstance(el, (list, tuple))] # filter out tuples and list
[1, 2, 3, 4, 5]

如何从python中的列表中删除所有列表？

2 个答案: