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}]

# 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

2014-05-15 00:07:38 +08:00

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

fdsfsdfsdf3334

2014-05-15 00:09:27 +08:00

好像有点问题，你看看能不能用

fdsfsdfsdf3334

2014-05-15 00:14:12 +08:00

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

2014-05-15 00:44:25 +08:00

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

2014-05-15 00:44:43 +08:00

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

kxxoling

2014-05-15 00:45:56 +08:00

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

2014-05-15 00:51:17 +08:00 via iPhone

这不是和女神合租的rubyer么，没认错吧

cbsw

2014-05-15 01:02:05 +08:00

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

2014-05-15 04:31:20 +08:00

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

2014-05-15 09:30:27 +08:00

```
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

2014-05-15 09:34:51 +08:00

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
```