Question

vowels=["a","e","i","o","u"]
word=["a","ir","the","book"]
for i in word:
    if len(i) == 1 and i in vowels:
        ....

我只是通过这个循环来检测'元音'数组中的'a'并且只有一个空格

喜欢'a'= 1个空格，'ant'= 3个空格。

我想将'a'与数组中的以下单词组合在一起。

如何将'a'与'ir'结合起来使'word'数组成为

word=["air","the","book"]

Answer 1

您可以使用迭代器构造新列表，只需添加LinearGradientBrush项，如果当前项为SliderTrackFillPressed，例如：

SliderTrackFillPointerOver

或者如果你碰巧需要添加多个元音：

<Page.Resources>
    <StaticResource x:Key="SliderTrackFillPressed" ResourceKey="MyLinearGradientBrush" />
    <StaticResource x:Key="SliderTrackFillPointerOver" ResourceKey="MyLinearGradientBrush" />
    <LinearGradientBrush x:Key="MyLinearGradientBrush" StartPoint="0.5,0" EndPoint="0.5,1">
        <GradientStop Color="#ff0000" Offset="0.0" />
        <GradientStop Color="#ffff00" Offset="0.2" />
        <GradientStop Color="#00ff00" Offset="0.4" />
        <GradientStop Color="#00ffff" Offset="0.6" />
        <GradientStop Color="#0000ff" Offset="0.8" />
        <GradientStop Color="#ff00ff" Offset="1.0" />
    </LinearGradientBrush>
</Page.Resources>
<Grid Background="{ThemeResource ApplicationPageBackgroundThemeBrush}">
    <Slider Grid.Column="1" Orientation="Vertical" Margin="6 0 0 0" Background="{StaticResource MyLinearGradientBrush}">
    </Slider>
</Grid>

Answer 2

>>> vowels = {'a', 'e', 'i', 'o', 'u'}  # used set for efficient `in` operation
>>> words = ['a', 'ir', 'the', 'book']
>>> for i in range(len(words) - 2, -1, -1):
...     if words[i] in vowels:
...         # len(words[i]) == 1 is redundant (all vowels items' length = 1)
...         words[i] += words.pop(i + 1)  # merge vowel with next word
... 
>>> words
['air', 'the', 'book']

注意：向后迭代;否则，只要list.pop调用，索引就会被破坏。

如何在一个数组中组合两个文本

2 个答案: