获取列表长度的一定百分比的索引
获取列表长度的一定百分比的索引
提问于 2015-04-05 16:57:28
def get_strings(letters, max_length):
for i in range(1, max_length + 1):
for value in product(letters, repeat=i):
yield "".join(value)
comb = [i for i in get_strings(ascii_lowercase, 4)]
print "# of possible combinations: %s" % len(comb)
def perc(i, tot):
p = float(i) /float(tot)
return p * 100
marker = [x for x in range(100) if x % 10 == 0]
marker.pop(0)
# make it 10:0, 20:0, 30:0, 40:0 and so forth...
mark = dict(zip(marker, [0 for i in range(len(marker))]))
print "BEFORE:"
for i in marker:
print "%s percent index: %s" % (i, mark[i])
l = len(comb)
for i,v in enumerate(comb):
p = perc(i, l)
ip = math.ceil(p)
if ip in marker:
mark[ip] = i
print "AFTER:"
for i in marker:
print "%s percent index: %s" % (i, mark[i])产出：
# of possible combinations: 475254
BEFORE:
10 percent index: 0
20 percent index: 0
30 percent index: 0
40 percent index: 0
50 percent index: 0
60 percent index: 0
70 percent index: 0
80 percent index: 0
90 percent index: 0
AFTER:
10 percent index: 47525
20 percent index: 95050
30 percent index: 142576
40 percent index: 190101
50 percent index: 237627
60 percent index: 285152
70 percent index: 332677
80 percent index: 380203
90 percent index: 427728我能够用上面的代码做这件事，但是它看起来很乏味，而且有很多不必要的步骤(或者可以组合或减少)。
有简化吗？
回答 2
Stack Overflow用户
回答已采纳
发布于 2015-04-05 17:08:57
让我们一分为二地解决这个问题。
首先，让我们简化百分比计算
perc = lambda i, t: (i * t) / 100现在让我们简化您的marker计算
marker = xrange(10, 100, 10)现在，让我们计算列表长度的某些百分比的索引：
for i in marker:
print '%s percent index: %s' % (i, perc(i, len(comb))就这样!
您可以将上面的内容简化为简洁的三行：
perc = lambda i, t: (i * t) / 100
for i in xrange(10, 100, 10):
print '%s percent index: %s' % (i, perc(i, len(comb))如果您确实需要将您的标记存储在mark dict中，请使用dict理解
mark = {i: perc(i, len(comb)) for i in xrange(10, 100, 10)}这部分代码可能是不必要的：
marker = [x for x in range(100) if x % 10 == 0]
marker.pop(0)
# make it 10:0, 20:0, 30:0, 40:0 and so forth...
mark = dict(zip(marker, [0 for i in range(len(marker))]))
print "BEFORE:"
for i in marker:
print "%s percent index: %s" % (i, mark[i])
l = len(comb)
for i,v in enumerate(comb):
p = perc(i, l)
ip = math.ceil(p)
if ip in marker:
mark[ip] = iStack Overflow用户
发布于 2015-04-05 17:20:22
您应该使用xrange，一些列表理解也是不必要的：
from collections import OrderedDict
def get_strings(letters, max_length):
return ("".join(value) for i in range(1, max_length + 1)
for value in product(letters, repeat=i))
comb = [i for i in get_strings(ascii_lowercase, 4)]
print "# of possible combinations: {}".format(len(comb))
# use an OrderedDict to maintain order, create the keys using a start
# of 10 and a step size of ten
mark = OrderedDict.fromkeys(xrange(10, 100, 10),0,)
# iterate over the items to avoid unnecessary lookups
print "BEFORE:"
for k, v in mark.iteritems():
print "{} percent index: {}".format(k, v)
l = len(comb)
# use keys of dict
for i in mark:
mark[i] = int(float(i) * l / 100)
print "AFTER:"
for k,v in mark.iteritems():
print "{} percent index: {}".format(k, v)不需要一个计算百分比的函数，除非您确实需要一个列表，否则range不应该在python2中使用。
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原文链接：
https://stackoverflow.com/questions/29459779
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