方法有点多啊 中规中矩的直接调用两次pop()
或者pop(3),pop(2)
或者pop(2),pop(2)
pop(-1),pop(-1)
pop(-2),pop(-1)
或者pop(3),pop(2)
或者pop(2),pop(2)
pop(-1),pop(-1)
pop(-2),pop(-1)
2016-03-28
score = 85
if score >= 90:
print 'excellent'
elif score >= 80:
print 'good'
elif score >= 60:
print 'passed'
else:
print 'failed'
if score >= 90:
print 'excellent'
elif score >= 80:
print 'good'
elif score >= 60:
print 'passed'
else:
print 'failed'
2016-03-28
t = ('a', 'b', ('A', 'B'))
print t
print t
2016-03-28
L = ['Adam', 'Lisa', 'Bart']
L[-1] = 'Adam'
L[-3] = 'Bart'
print L
L[-1] = 'Adam'
L[-3] = 'Bart'
print L
2016-03-28
L = [95.5,85,59]
print L[0];
print L[1];
print L[2];
print L;http://www.imooc.com/code/3357
print L[0];
print L[1];
print L[2];
print L;http://www.imooc.com/code/3357
2016-03-28
L = ['Adam', 95.5, 'Lisa', 85, 'Bart', 59]
print L
print L
2016-03-28
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
2016-03-28
# -*- coding: utf-8 -*-
print '''静夜思
床前明月光,
疑是地上霜。
举头望明月。
低头思故乡。'''
print '''静夜思
床前明月光,
疑是地上霜。
举头望明月。
低头思故乡。'''
2016-03-28
d = { 'Adam': 95, 'Lisa': 85, 'Bart': 59, 'Paul': 74 }
sum = 0.0
for k, v in d.items():
sum = sum + v
print k,':',v
print 'average', ':', 1.0*sum/len(d)
sum = 0.0
for k, v in d.items():
sum = sum + v
print k,':',v
print 'average', ':', 1.0*sum/len(d)
2016-03-28
d = { 'Adam': 95, 'Lisa': 85, 'Bart': 59, 'Paul': 74 }
sum = 0.0
for count in d.itervalues():
sum = sum + count
print 1.0*sum/len(d)
sum = 0.0
for count in d.itervalues():
sum = sum + count
print 1.0*sum/len(d)
2016-03-28