Python 重新排列字典，使值成为下一个键

我有一本字典如下：

{1: 3, 2: 4, 3: 5, 4: 1, 5: 2}

我想重新排列它，使当前对的值成为下一对的键，如下所示：

{1: 3, 3: 5, 5: 2, 2: 4, 4: 1}

---

如何实现这一点？

使用以下代码可以实现这一点-自（Python 3.6 for CPython）起的字典是按插入顺序排列的，因此我们从

d

中的第一个键开始，然后迭代，更新下一个键以使用最后一个键的值进行检查，直到我们新字典的长度和旧字典一样

代码：

d = {1: 3, 2: 4, 3: 5, 4: 1, 5: 2}

curr = list(d.keys())[0]

d_new = {}

while len(d) > len(d_new):
    d_new[curr] = curr = d[curr]

>>> d_new
{1: 3, 3: 5, 5: 2, 2: 4, 4: 1}

from collections import OrderedDict

d = {1: 3, 2: 4, 3: 5, 4: 1, 5: 2}

curr = 1

d_new = OrderedDict()

while len(d) > len(d_new):
    d_new[curr] = curr = d[curr]

d = {1: 3, 2: 4, 3: 1, 4: 1, 5: 2}

curr = list(d.keys())[0]

d_new = {}

while len(d) > len(d_new):
    if curr not in d_new:
        d_new[curr] = curr = d[curr]
    else:
        curr = next(x for x in d.keys() if x not in d_new)

>>> d_new
{1: 3, 3: 1, 2: 4, 4: 1, 5: 2}

输出：

d = {1: 3, 2: 4, 3: 5, 4: 1, 5: 2}

curr = list(d.keys())[0]

d_new = {}

while len(d) > len(d_new):
    d_new[curr] = curr = d[curr]

>>> d_new
{1: 3, 3: 5, 5: 2, 2: 4, 4: 1}

from collections import OrderedDict

d = {1: 3, 2: 4, 3: 5, 4: 1, 5: 2}

curr = 1

d_new = OrderedDict()

while len(d) > len(d_new):
    d_new[curr] = curr = d[curr]

d = {1: 3, 2: 4, 3: 1, 4: 1, 5: 2}

curr = list(d.keys())[0]

d_new = {}

while len(d) > len(d_new):
    if curr not in d_new:
        d_new[curr] = curr = d[curr]
    else:
        curr = next(x for x in d.keys() if x not in d_new)

>>> d_new
{1: 3, 3: 1, 2: 4, 4: 1, 5: 2}

在Python 3.7下面，您可以使用：

d = {1: 3, 2: 4, 3: 5, 4: 1, 5: 2}

curr = list(d.keys())[0]

d_new = {}

while len(d) > len(d_new):
    d_new[curr] = curr = d[curr]

>>> d_new
{1: 3, 3: 5, 5: 2, 2: 4, 4: 1}

from collections import OrderedDict

d = {1: 3, 2: 4, 3: 5, 4: 1, 5: 2}

curr = 1

d_new = OrderedDict()

while len(d) > len(d_new):
    d_new[curr] = curr = d[curr]

d = {1: 3, 2: 4, 3: 1, 4: 1, 5: 2}

curr = list(d.keys())[0]

d_new = {}

while len(d) > len(d_new):
    if curr not in d_new:
        d_new[curr] = curr = d[curr]
    else:
        curr = next(x for x in d.keys() if x not in d_new)

>>> d_new
{1: 3, 3: 1, 2: 4, 4: 1, 5: 2}

根据下面的注释，如果键/值包含循环，我们希望移动到下一个键-值对。这可以通过以下方式实现：

{1: 3, 2: 4, 3: 5, 4: 1, 5: 2}

代码：

d = {1: 3, 2: 4, 3: 5, 4: 1, 5: 2}

curr = list(d.keys())[0]

d_new = {}

while len(d) > len(d_new):
    d_new[curr] = curr = d[curr]

>>> d_new
{1: 3, 3: 5, 5: 2, 2: 4, 4: 1}

from collections import OrderedDict

d = {1: 3, 2: 4, 3: 5, 4: 1, 5: 2}

curr = 1

d_new = OrderedDict()

while len(d) > len(d_new):
    d_new[curr] = curr = d[curr]

d = {1: 3, 2: 4, 3: 1, 4: 1, 5: 2}

curr = list(d.keys())[0]

d_new = {}

while len(d) > len(d_new):
    if curr not in d_new:
        d_new[curr] = curr = d[curr]
    else:
        curr = next(x for x in d.keys() if x not in d_new)

>>> d_new
{1: 3, 3: 1, 2: 4, 4: 1, 5: 2}

输出：

d = {1: 3, 2: 4, 3: 5, 4: 1, 5: 2}

curr = list(d.keys())[0]

d_new = {}

while len(d) > len(d_new):
    d_new[curr] = curr = d[curr]

>>> d_new
{1: 3, 3: 5, 5: 2, 2: 4, 4: 1}

from collections import OrderedDict

d = {1: 3, 2: 4, 3: 5, 4: 1, 5: 2}

curr = 1

d_new = OrderedDict()

while len(d) > len(d_new):
    d_new[curr] = curr = d[curr]

d = {1: 3, 2: 4, 3: 1, 4: 1, 5: 2}

curr = list(d.keys())[0]

d_new = {}

while len(d) > len(d_new):
    if curr not in d_new:
        d_new[curr] = curr = d[curr]
    else:
        curr = next(x for x in d.keys() if x not in d_new)

>>> d_new
{1: 3, 3: 1, 2: 4, 4: 1, 5: 2}

---

您可以使用itertools中的累积来完成此操作：

d = {1: 3, 2: 4, 3: 5, 4: 1, 5: 2}

from itertools import accumulate
od = {k:d[k] for k in accumulate(d,lambda k,_:d[k]) }

#od : {1: 3, 3: 5, 5: 2, 2: 4, 4: 1}

---

就我所记得的，python dict没有指定要排序，因此在键值对中没有排序。打印词典时看到的顺序是任意的。编辑：似乎是这样的：）我认为您必须创建一个新的字典，并从一开始就按正确的顺序插入键，因为我认为它们不可重新排序。查看OrderedDict结构类型：检查实际上，从Python 3.6开始，

{k:d[k]for k in accumulate（d，d.get）}

也可以工作。