Python 如何实现，数字 1 到 10，如何通过程序实现[1:{2:3, 4:5},6:{7:8, 9:10}]

[zedde]

如题，数字1到10，如何通过程序实现[1:{2:3, 4:5},6:{7:8, 9:10}]
不胜感激~

上面显示结果有误，比如数字1到20，实现结果
{1:{2:3, 4:5}, 6:{7:8, 9:10}, 11:{12:13, 14:15}, 16:{17:18, 19:20}}

[Muninn]

不太理解

[zedde]

@Muninn 应该是{1:{2:3,4:5},6:{7:8,9:10}} 嵌套字典。

[zedde]

1-20 类似{1:{2:3, 4:5}, 6:{7:8, 9:10}, 11:{12:13, 14:15}, 16:{17:18, 19:20}}

[fdsfsdfsdf3334]

# coding=utf-8

t={} #初始化字典

for x in range(1,21):
if x%5==1:
print x
t[x]="{"+str(x+1)+":"+str(x+2)+","+str(x+3)+":"+str(x+4)+"}"




t1= sorted(t.iteritems(), key=lambda d:d[0], reverse = False) #按照键值排序

print t1

[fdsfsdfsdf3334]

http://paste.ubuntu.com/7463487/

[fdsfsdfsdf3334]

好像有点问题，你看看能不能用

[fdsfsdfsdf3334]

print "---------------------------------------------\n\r"
print "t:\n\r"
print t
print t[1]

print "---------------------------------------------\n\r"
print "t1:\n\r"
print t1
print t1[0][1]

[kxxoling]

我理解你是想通过一个list获取这样的dict，list的长度要能被5整除。
```
l = range(1, 21)
assert len(l)%5 == 0
d = {}
for i in range(0, len(l), 5):
d[l[i]] = {l[i+1]: l[i+2], l[i+3]: l[i+4]}
print d
```

[delo]

http://gist.github.com/2495bd648c72f79a426a

[kxxoling]

额，v2ex不支持这么加代码。。。不过逻辑不复杂，我就不用gist了，讲究看看吧～

[palytoxin]

这不是和女神合租的rubyer么，没认错吧

[cbsw]

一行代码而已
{k:{k+1:k+2, k+3:k+4} for k in [5*i+1 for i in range(4)]}
如果要写成函数，就把最后一个 4 作为参数

[kavinyao]

In [13]: {i: {j:j+1 for j in [i+1, i+3]} for i in range(1, 20, 5)}
Out[13]: {1: {2: 3, 4: 5}, 6: {7: 8, 9: 10}, 11: {12: 13, 14: 15}, 16: {17: 18, 19: 20}}

[daniel7725]

```
def test(num):
if num <5:
return None
targe = {}
for i in xrange(1,num,5):
tmp_dict = {str(i):{str(i+1):str(i+2),str(i+3):str(i+4)}}
targe.update(tmp_dict)
return targe

[daniel7725]

看看v2ex支不支持markdown
```
def test(num):
if num <5:
return None
targe = {}
for i in xrange(1,num,5):
tmp_dict = {str(i):{str(i+1):str(i+2),str(i+3):str(i+4)}}
targe.update(tmp_dict)
return targe
```

[wangyongbo]

{ x : { x + 1 :x+ 2, x+ 3: x+ 4} for x in range(1, 20, 5)}

[wangyongbo]

{ x : { i : i + 1 for i in xrange(x+1, x+4, 2)} for x in range(1, 20, 5)}

[toctan]

Ruby:

(1..20).step(5).reduce({}) { |r, x| r.merge(x => Hash[*(x + 1..x + 4)]) }

or the lazy way:

Hash.new { |h, k| h[k] = Hash[*(k + 1..k + 4)]) }

[hanks315]

https://gist.github.com/hanks/e3b1bf128209c868d910

[foomorrow]

n = 2
dict(map(lambda x:(x,dict(((x+1,x+2),(x+3,x+4)))),[x+1 for x in xrange(0,n*10,5)]))