Question

我想使用pythons map计算字符串中的字母频率，并减少内置函数。任何人都可以提供一些有关我如何做到这一点的见解吗？

到目前为止我得到了什么：

s = "the quick brown fox jumped over the lazy dog"

# Map function
m = lambda x: (x,1)

# Reduce
# Add the two frequencies if they are the same
# else.... Not sure how to put both back in the list
# in the case where they are not the same.
r = lambda x,y: (x[0], x[1] + y[1]) if x[0] == y[0] else ????

freq = reduce(r, map(m, s))

当所有字母都相同时，这很有效。

>>> s
'aaaaaaa'
>>> map(m, s)
[('a', 1), ('a', 1), ('a', 1), ('a', 1), ('a', 1), ('a', 1), ('a', 1)]
>>> reduce(r, map(m, s))
('a', 7)

当有不同的字母时，如何让它很好地工作？

Answer 1

稍等一下关于代码的问题，我将指出通常（和最快）计算事物的方法之一是使用collections模块中的Counter类。以下是Python 2.7.3解释器中使用的示例：

>>> from collections import Counter
>>> lets=Counter('aaaaabadfasdfasdfafsdff')
>>> lets
Counter({'a': 9, 'f': 6, 'd': 4, 's': 3, 'b': 1})
>>> s = "the quick brown fox jumped over the lazy dog"
>>> Counter(s)
Counter({' ': 8, 'e': 4, 'o': 4, 'd': 2, 'h': 2, 'r': 2, 'u': 2, 't': 2, 'a': 1, 'c': 1, 'b': 1, 'g': 1, 'f': 1, 'i': 1, 'k': 1, 'j': 1, 'm': 1, 'l': 1, 'n': 1, 'q': 1, 'p': 1, 'w': 1, 'v': 1, 'y': 1, 'x': 1, 'z': 1})

要使用reduce，请定义一个辅助函数addto(oldtotal,newitem)，将newitem添加到oldtotal并返回新的总计。 total的初始值设定项是空字典{}。这是一个解释的例子。请注意，get（）的第二个参数是当键尚未出现在字典中时使用的默认值。

 >>> def addto(d,x):
...     d[x] = d.get(x,0) + 1
...     return d
... 
>>> reduce (addto, s, {})
{' ': 8, 'a': 1, 'c': 1, 'b': 1, 'e': 4, 'd': 2, 'g': 1, 'f': 1, 'i': 1, 'h': 2, 'k': 1, 'j': 1, 'm': 1, 'l': 1, 'o': 4, 'n': 1, 'q': 1, 'p': 1, 'r': 2, 'u': 2, 't': 2, 'w': 1, 'v': 1, 'y': 1, 'x': 1, 'z': 1}

下面显示的代码打印了几种方法的1000次传递的执行时间。在旧的AMD Athlon 5000+ Linux 3.2.0-32 Ubuntu 12系统上执行时，打印出两个不同的字符串s：

String length is 44   Pass count is 1000
horsch1 : 0.77517914772
horsch2 : 0.778718948364
jreduce : 0.0403778553009
jcounter: 0.0699260234833
String length is 4931   Pass count is 100
horsch1 : 8.25176692009
horsch2 : 8.14318394661
jreduce : 0.260674953461
jcounter: 0.282369852066

（reduce方法的运行速度比Counter方法略快。）时序码如下。它使用timeit模块。在这里的代码中，timeit.Timer的第一个参数是要重复计时的代码，第二个参数是设置代码。

import timeit
from collections import Counter
passes = 1000

m1 = lambda x: [int(ord(x) == i) for i in xrange(65,91)]

def m2(x):
    return [int(ord(x) == i) for i in xrange(65,91)]

def es1(s):
    add = lambda x,y: [x[i]+y[i] for i in xrange(len(x))]
    freq = reduce(add,map(m1, s.upper()))
    return freq

def es2(s):
    add = lambda x,y: [x[i]+y[i] for i in xrange(len(x))]
    freq = reduce(add,map(m2, s.upper()))
    return freq

def addto(d,x):
    d[x] = d.get(x,0) + 1
    return d

def jwc(s):
    return Counter(s)

def jwr(s):
    return reduce (addto, s, {})

s = "the quick brown fox jumped over the lazy dog"
print 'String length is',len(s), '  Pass count is',passes
print "horsch1 :",timeit.Timer('f(s)', 'from __main__ import s, m1,     es1 as f').timeit(passes)
print "horsch2 :",timeit.Timer('f(s)', 'from __main__ import s, m2,     es2 as f').timeit(passes)
print "jreduce :",timeit.Timer('f(s)', 'from __main__ import s, addto,  jwr as f').timeit(passes)
print "jcounter:",timeit.Timer('f(s)', 'from __main__ import s, Counter,jwc as f').timeit(passes)

Answer 2

ord()通常会给出ascii数字。我的方法计算字母的频率，其中每个索引对应于字母表中该位置的字母。由于您对字符串进行了上限，因此该方法不区分大小写。

s = "the quick brown fox jumped over the lazy dog"

# Map function
m = lambda x: [ord(x) == i for i in xrange(0,26)]

add = lambda x,y: [x[i]+y[i] for i in xrange(len(x))]
freq = reduce(add,map(m, s.upper()))

Answer 3

您还可以使用defaultdict：

>>> from collections import defaultdict
>>> d = defaultdict(int)
>>> s = 'the quick brown fox jumped over the lazy dog'
>>> for i in s:
...    d[i] += 1
...
>>> for letter,count in d.iteritems():
...    print letter,count
...
  8 # number of spaces
a 1
c 1
b 1
e 4
d 2
g 1
f 1
i 1
h 2
k 1
j 1
m 1
l 1
o 4
n 1
q 1
p 1
r 2
u 2
t 2
w 1
v 1
y 1
x 1
z 1

Answer 4

您还可以使用s.count方法：

{x: s.count(x) for x in set(s)}

请注意，我使用set(s)只计算一次字符串中每个字母的频率。这是我的机器上的测试结果：

String length is 44   Pass count is 1000
horsch1  : 0.317646980286
horsch2  : 0.325616121292
jreduce  : 0.0106990337372
jcounter : 0.0142340660095
def_dict : 0.00750803947449
just_dict: 0.00737881660461
s_count  : 0.00887513160706

String length is 4400   Pass count is 100
horsch1  : 3.24123382568
horsch2  : 3.23079895973
jreduce  : 0.0944828987122
jcounter : 0.102299928665
def_dict : 0.0341360569
just_dict: 0.0643239021301
s_count  : 0.0224709510803

这是一个测试代码：

import timeit
from collections import Counter, defaultdict
passes = 100

m1 = lambda x: [int(ord(x) == i) for i in xrange(65,91)]

def m2(x):
    return [int(ord(x) == i) for i in xrange(65,91)]

def es1(s):
    add = lambda x,y: [x[i]+y[i] for i in xrange(len(x))]
    freq = reduce(add,map(m1, s.upper()))
    return freq

def es2(s):
    add = lambda x,y: [x[i]+y[i] for i in xrange(len(x))]
    freq = reduce(add,map(m2, s.upper()))
    return freq

def addto(d,x):
    d[x] = d.get(x,0) + 1
    return d

def jwc(s):
    return Counter(s)

def jwr(s):
    return reduce (addto, s, {})

def def_dict(s):
    d = defaultdict(int)
    for i in s:
        d[i]+=1
    return d

def just_dict(s):
    freq = {}
    for i in s:
        freq[i]=freq.get(i, 0) + 1
    return freq

def s_count(s):
    return {x: s.count(x) for x in set(s)}

s = "the quick brown fox jumped over the lazy dog"*100
print 'String length is',len(s), '  Pass count is',passes
print "horsch1  :",timeit.Timer('f(s)', 'from __main__ import s, m1,     es1 as f').timeit(passes)
print "horsch2  :",timeit.Timer('f(s)', 'from __main__ import s, m2,     es2 as f').timeit(passes)
print "jreduce  :",timeit.Timer('f(s)', 'from __main__ import s, addto,  jwr as f').timeit(passes)
print "jcounter :",timeit.Timer('f(s)', 'from __main__ import s, Counter,jwc as f').timeit(passes)
print "def_dict :",timeit.Timer('f(s)', 'from __main__ import s, defaultdict, def_dict as f').timeit(passes)
print "just_dict:",timeit.Timer('f(s)', 'from __main__ import s, just_dict as f').timeit(passes)
print "s_count  :",timeit.Timer('f(s)', 'from __main__ import s, s_count as f').timeit(passes)

如何使用pythons内置map和reduce函数计算字符串中的字母频率

4 个答案: