s = set(['Adam',' Lisa',' Bart',' Paul'])
2016-04-26
d = {
'Adam': 95,
'Lisa': 85,
'Bart': 59
}
for key in d:
print key,':',d[key]
'Adam': 95,
'Lisa': 85,
'Bart': 59
}
for key in d:
print key,':',d[key]
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d = {
95: 'Adam',
85: 'Lisa',
59: 'Bart'
}
d[72]='Paul'
95: 'Adam',
85: 'Lisa',
59: 'Bart'
}
d[72]='Paul'
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# -*- coding: utf-8 -*-
d = {
95:'Adam',
85: 'Lisa',
59: 'Bart'
}
d = {
95:'Adam',
85: 'Lisa',
59: 'Bart'
}
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d = {
'Adam': 95,
'Lisa': 85,
'Bart': 59
}
print 'Adam',':',d['Adam']
print 'Lisa',':',d['Lisa']
print 'Bart',':',d['Bart']
'Adam': 95,
'Lisa': 85,
'Bart': 59
}
print 'Adam',':',d['Adam']
print 'Lisa',':',d['Lisa']
print 'Bart',':',d['Bart']
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d = {
'Adam': 95,
'Lisa': 85,
'Bart': 59,
'Paul': 75
}
'Adam': 95,
'Lisa': 85,
'Bart': 59,
'Paul': 75
}
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for x in ['1','2','3','4','5','6','7','8']:
for y in ['2','3','4','5','6','7','8','9']:
if x<y:
print x+y
for y in ['2','3','4','5','6','7','8','9']:
if x<y:
print x+y
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sum = 0
x = 0
while True:
x = x + 1
if x > 100:
break
if x%2==0:
continue
sum+= x
print sum
x = 0
while True:
x = x + 1
if x > 100:
break
if x%2==0:
continue
sum+= x
print sum
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sum = 0
x = 1
n = 1
while True:
sum+=x
x*=2
n+=1
if n>20:
break
print sum
x = 1
n = 1
while True:
sum+=x
x*=2
n+=1
if n>20:
break
print sum
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L = [75, 92, 59, 68]
sum = 0.0
for a in L:
sum+=a
print sum / 4
sum = 0.0
for a in L:
sum+=a
print sum / 4
2016-04-26
已采纳回答 / 清波
<...code...>首先, *args 接受不定数量的参数, 在函数中调用的时候,不用星号(*)了, 在函数中调用的时候 args 就是传入的参数 组成的 元组(tuple).其次, 有了 len(args) , 就不用另外定义i 并且自加1 了。 最后,为了防止 发生 除0 的错误,进入函数后,第一时间判断参数长度。所以, 根据上面三条,理解下正确答案。
2016-04-26