Question

我有两个列表，其值与其他位置匹配。因此Jon的得分为123，而Bede的成绩为11等。

name = ["Jon", "Bede", "Joe"]
score = [123, 11, 43]

如何订购列表，以便输出最低分，首先提升到最高分，每次输出得分者的姓名。

Answer 1

for s, n in sorted(zip(score, name)): # sort by score.
    print(n, s)

Answer 2

如果您有两个列表，其中索引是相关的，则很难使它们保持同步。解决方案python有一个元组列表，其中每个元组中的第一个元素是名称，第二个元素是分数：

names_with_score = [
    ("Jon", 123),
    ("Bede", 11),
    ("Joe", 43),
]
names_with_score.sort(key=lambda x: x[1]) # sort with the second element

如果您无法控制数据的传送方式，您可以使用zip-function加入列表：

names_with_score = zip(name, score)

Answer 3

这个怎么样？

Bede: 11
Joe: 43
Jon: 123

输出：

        public boolean onDependentViewChanged(CoordinatorLayout parent, View child,
            View dependency) {
        final CoordinatorLayout.Behavior behavior =
                ((CoordinatorLayout.LayoutParams) dependency.getLayoutParams()).getBehavior();
        if (behavior instanceof Behavior) {
            // Offset the child so that it is below the app-bar (with any overlap)

            final int appBarOffset = ((Behavior) behavior)
                    .getTopBottomOffsetForScrollingSibling();
            final int expandedMax = dependency.getHeight() - mOverlayTop;
            final int collapsedMin = parent.getHeight() - child.getHeight();

            if (mOverlayTop != 0 && dependency instanceof AppBarLayout) {
                // If we have an overlap top, and the dependency is an AppBarLayout, we control
                // the offset ourselves based on the appbar's scroll progress. This is so that
                // the scroll happens sequentially rather than linearly
                final int scrollRange = ((AppBarLayout) dependency).getTotalScrollRange();
                setTopAndBottomOffset(AnimationUtils.lerp(expandedMax, collapsedMin,
                        Math.abs(appBarOffset) / (float) scrollRange));
            } else {
                setTopAndBottomOffset(dependency.getHeight() - mOverlayTop + appBarOffset);
            }
        }
        return false;
    }

Answer 4

如果你不介意的话，这会合并列表：

sorted(zip(name,score))

Answer 5

尝试将它们转换为字典以便于访问，而不是坚持使用一对并行的list。然后，您可以使用key参数按键对字典进行排序：

name = ["Jon", "Bede", "Joe"]
score = [123, 11, 43]
results = dict(zip(name, score))
for k in sorted(results, key=results.get):
    print('{}: {}'.format(k, results[k]))

结果：

Bede: 11
Joe: 43
Jon: 123

在我对其中一个数组进行排序后，如何对齐两个数组？

5 个答案: