d = {
'Adam': 95,
'Lisa': 85,
'Bart': 59
}
print 'Adam:',d.get('Adam')
print 'Lisa:',d.get('Lisa')
print 'Bart:',d.get('Bart')
'Adam': 95,
'Lisa': 85,
'Bart': 59
}
print 'Adam:',d.get('Adam')
print 'Lisa:',d.get('Lisa')
print 'Bart:',d.get('Bart')
2016-07-11
教大家一个既简单又简洁的
for x in range(1,9):
for y in range(x+1,10):
print x*10+y
for x in range(1,9):
for y in range(x+1,10):
print x*10+y
2016-07-11
注意:x必须先赋值为0,而不是1
sum = 0
x = 0
while True:
x = x + 1
if x > 100:
break
elif x%2==0:
continus
sum+=x
print sum
sum = 0
x = 0
while True:
x = x + 1
if x > 100:
break
elif x%2==0:
continus
sum+=x
print sum
2016-07-11
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
2016-07-11
学过C语言的话,这个理解起来很容易,学计算机首先要掌握的第一门语言是C语言,觉得吃力的同学可以先学C,然后学python就很容易了,学到目前为止,感觉python就是c语言的简化版
L = [75, 92, 59, 68]
sum=0.0
for name in L:
sum=sum+name
print sum/4 输出结果为73.5
L = [75, 92, 59, 68]
sum=0.0
for name in L:
sum=sum+name
print sum/4 输出结果为73.5
2016-07-10
看评论无语了都,看我的多简单
L = ['Adam', 'Lisa', 'Bart']
L[0]='Bart'
L[2]='Adam'
print L
L = ['Adam', 'Lisa', 'Bart']
L[0]='Bart'
L[2]='Adam'
print L
2016-07-10
已采纳回答 / Dai9210
运用短路法则来判断,逐个运行,注意逻辑运算的优先顺序,and优先级高于or。举2个例子:True and False or Ture # ==> 先运算True and False 结果为F,再运算后面的F or Ture ,整体结果为T>>> print 2%2==0 and 3%2 == 0 or 4%2 ==0 True or False and False # ==> 先运算False and False 结果为F,再运算后面的Ture or F,整体结果为T&...
2016-07-10