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', ':', sum/len(d.items())
sum = 0.0
for k, v in d.items():
sum = sum + v
print k,':',v
print 'average', ':', sum/len(d.items())
2016-03-02
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', ':', sum/len(d)
sum = 0.0
for k, v in d.items():
sum = sum + v
print k,':',v
print 'average', ':', sum/len(d)
2016-03-02
d = { 'Adam': 95, 'Lisa': 85, 'Bart': 59, 'Paul': 74 }
sum = 0.0
i=0
for f in d.values():
sum =sum +f
i=i+1
print sum/i
d = { 'Adam': 95, 'Lisa': 85, 'Bart': 59, 'Paul': 74 }
sum = 0.0
for f in d.values():
sum =sum +f
print sum/len(d.values())
sum = 0.0
i=0
for f in d.values():
sum =sum +f
i=i+1
print sum/i
d = { 'Adam': 95, 'Lisa': 85, 'Bart': 59, 'Paul': 74 }
sum = 0.0
for f in d.values():
sum =sum +f
print sum/len(d.values())
2016-03-02
假设有如下的dict:
d = { 'Adam': 95, 'Lisa': 85, 'Bart': 59 }
完全可以通过一个复杂的列表生成式把它变成一个 HTML 表格:
tds = ['<tr><td>%s</td><td>%s</td></tr>' % (name, score) for name, score in d.iteritems()]
print '<table>'
print '<tr><th>Name</th><th>Score</th><tr>'
print '\n'.join(tds)
print '</table>'
d = { 'Adam': 95, 'Lisa': 85, 'Bart': 59 }
完全可以通过一个复杂的列表生成式把它变成一个 HTML 表格:
tds = ['<tr><td>%s</td><td>%s</td></tr>' % (name, score) for name, score in d.iteritems()]
print '<table>'
print '<tr><th>Name</th><th>Score</th><tr>'
print '\n'.join(tds)
print '</table>'
2016-03-02
最赞回答 / lc云泽
可以用负数,表示倒数,但用负数时要注意的是,新元素添加到原数列的指定位置,并把原数列指定位置的元素往后推一位<...code...>4这个元素把3的位置占了,并把3往后推了一位
2016-03-02
def square_of_sum(L):
sum = 0
for a in L:
t = a * a
sum = sum + t
return sum
print square_of_sum([1, 2, 3, 4, 5])
print square_of_sum([-5, 0, 5, 15, 25])
sum = 0
for a in L:
t = a * a
sum = sum + t
return sum
print square_of_sum([1, 2, 3, 4, 5])
print square_of_sum([-5, 0, 5, 15, 25])
2016-03-02