Question

我有一个看起来像下面的列表

[('Jon', '177'), ('Danei', '1420'), ('Adan', '146'), ('Gabriel', '110'), ('Chase', '117'), ('Hetmyer', '418'), ('Smith', '162')]

这实际上是一个自定义排序列表，我想通过固定列表中的第一个元素来更改其顺序，即如果我将Hetmyer作为输入，新列表应该是这样

[('Hetmyer', '418'), ('Smith', '162'),('Jon', '177'), ('Danei', '1420'), ('Adan', '146'), ('Gabriel', '110'), ('Chase', '117')]

如果我将Adan作为输入，新列表应如下所示

[('Adan', '146'), ('Gabriel', '110'), ('Chase', '117'), ('Hetmyer', '418'), ('Smith', '162'),('Jon', '177'), ('Danei', '1420') ]

基本上，我给出的任何输入都应成为列表的第一个成员，而其余顺序应保持原样，即它应从输入开始，并在输入成员之前以列表成员结束，我们有任何内置函数吗

Answer 1

找到位置

for i, item in enumerate(l):
    if item[0] == "Hetmyer":
        pos = i

建立新列表：

m = l[pos:] + l[:pos]

结果：

[('Hetmyer', '418'), ('Smith', '162'), ('Jon', '177'), ('Danei', '1420'), ('Adan', '146'), ('Gabriel', '110'), ('Chase', '117')]

Answer 2

s = input()

 a = [('Jon', '177'), ('Danei', '1420'), ('Adan', '146'), ('Gabriel', '110'), ('Chase', '117'), ('Hetmyer', '418'), ('Smith', '162')]


for i in a:
    if i[0] == s:
        a = a[a.index(i):] + a[:a.index(i)]

print(a) #[('Adan', '146'), ('Gabriel', '110'), ('Chase', '117'), ('Hetmyer', '418'), ('Smith', '162'), ('Jon', '177'), ('Danei', '1420')]

Answer 3

您可以继续使用zip使用index查找元素，然后进行切片：

def resort(name_list, name):
    """Return the given name_list ordered using the given name."""
    index = list(zip(*name_list))[0].index(name)
    return name_list[:index] + name_list[index:]

resort(names, "Danei")

结果：

[('Jon', '177'),
 ('Danei', '1420'),
 ('Adan', '146'),
 ('Gabriel', '110'),
 ('Chase', '117'),
 ('Hetmyer', '418'),
 ('Smith', '162')]

根据输入更改列表的顺序

3 个答案: