-
-
def mySort(a):
-
lo=len(a)-1
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i=0
-
while i<lo+1:
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j=lo
-
while j>=i:
-
if(a[j-1]>a[j]):
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a[j],a[j-1]=a[j-1],a[j]
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j=j-1
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i=i+1
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if __name__ == '__main__':
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l1=[3,2]
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mySort(l1)
- print l1
python 字典排序
-
xychong123
2017-03-17 11:59:01 -
IT综合
- 原创
字典排序：
d1={'y': 2, 'x': 1, 'z': 3,'a':99,'c':23,'f':15}
字典转换序列:
c1=d1.items()
print c1
[('a', 99), ('c', 23), ('f', 15), ('x', 1), ('y', 2), ('z', 3)]
key,values位置互换：
c2=[]
for i in c1:
c2.append((i[1],i[0]))
In [15]: print c2
[(99, 'a'), (23, 'c'), (15, 'f'), (2, 'y'), (1, 'x'), (3, 'z')]
重新生成字典：
d2={}
for i in c2:
d2[i[0]]=i[1]
In [16]: print d2
{1: 'x', 2: 'y', 99: 'a', 15: 'f', 3: 'z', 23: 'c'}
进行排序：
for i in sorted(d2.keys()):
print d2[i],':',i
x : 1
y : 2
z : 3
f : 15
c : 23
a : 99