Question

这是对早期问题的后续跟进。我得到了一些很好的建议，所以我想我会再试一次运气。

from itertools import takewhile

if K is None:
    illuminacond = lambda x: x.split(',')[0] != '[Controls]'
else:
    illuminacond = lambda x: x.split(',')[0] != '[Controls]' and i < K

af=open('a')
bf=open('b', 'w')
cf=open('c', 'w')

i = 0
if K is None:
    for line in takewhile(illuminacond, af):
        line_split=line.split(',')
        pid=line_split[1][0:3]
        out = line_split[1] + ',' + line_split[2] + ',' + line_split[3][1] + line_split[3][3] + ',' \
                                  + line_split[15] + ',' + line_split[9] + ',' + line_split[10]
        if pid!='cnv' and pid!='hCV' and pid!='cnv':
            i = i+1
            bf.write(out.strip('"')+'\n')
            cf.write(line)
else:
    for line in takewhile(illuminacond, af):
        line_split=line.split(',')
        pid=line_split[1][0:3]
        out = line_split[1] + ',' + line_split[2] + ',' + line_split[3][1] + line_split[3][3] + ',' \
                            + line_split[15] + ',' + line_split[9] + ',' + line_split[10]
        if pid!='cnv' and pid!='hCV' and pid!='cnv':
            i = i+1
            bf.write(out.strip('"')+'\n')

是否有可能使此代码变得紧凑？如果我在这样的两个循环中有一些共同点，一个明显的可能性就是将公共代码分解出来，但是在这里，eww。令人讨厌的是，这里唯一的区别是写给c。

代码简要摘要：如果K不是无，则循环K行a并写入b和c。否则，遍历所有a，只需写入b。

Answer 1

为什么不只使用一个循环，但包含条件在循环中？另外，我认为你可以摆脱lambda中的冗余。

from itertools import takewhile

k_is_none = K is None

def illuminacond(x):
    global i
    global K
    result = x.split(',')[0] != '[Controls]'
    if not k_is_none:
        result = result and i < K
    return result

af=open('a')
bf=open('b', 'w')
cf=open('c', 'w')

i = 0
for line in takewhile(illuminacond, af):
    line_split=line.split(',')
    pid=line_split[1][0:3]
    out = line_split[1] + ',' + line_split[2] + ',' + line_split[3][1] + line_split[3][3] + ',' \
                              + line_split[15] + ',' + line_split[9] + ',' + line_split[10]
    if pid!='cnv' and pid!='hCV' and pid!='cnv':
        i = i+1
        bf.write(out.strip('"')+'\n')
        if k_is_none:
            cf.write(line)

Answer 2

一次检查，一次循环，没有类，psyco可优化。

from itertools import takewhile

if K is None:
    illuminacond = lambda x: x.split(',')[0] != '[Controls]'
    def action(cf, line): cf.write(line)
else:
    illuminacond = lambda x: x.split(',')[0] != '[Controls]' and i < K
    def action(cf, line): pass

af=open('a')
bf=open('b', 'w')
cf=open('c', 'w')

i = 0
for line in takewhile(illuminacond, af):
    line_split=line.split(',')
    pid=line_split[1][0:3]
    out = line_split[1] + ',' + line_split[2] + ',' + line_split[3][1] + line_split[3][3] + ',' \
                              + line_split[15] + ',' + line_split[9] + ',' + line_split[10]
    if pid!='cnv' and pid!='hCV' and pid!='cnv':
        i = i+1
        bf.write(out.strip('"')+'\n')
        action(cf, line)

Answer 3

为什么不呢：

from itertools import takewhile

illuminacond = lambda x: x.split(',')[0] != '[Controls]' and (K is None or i<K) #i'm not so sure about this part, confused me a little :).

af=open('a')
bf=open('b', 'w')
cf=open('c', 'w')

for line in takewhile(illuminacond, af):
    line_split=line.split(',')
    pid=line_split[1][0:3]
    out = line_split[1] + ',' + line_split[2] + ',' + line_split[3][1] + line_split[3][3] + ',' \
                              + line_split[15] + ',' + line_split[9] + ',' + line_split[10]
    if pid!='cnv' and pid!='hCV' and pid!='cnv':
        i = i+1
        bf.write(out.strip('"')+'\n')
        if K is None:
            cf.write(line)

Answer 4

这个（基于第二类的版本）怎么样？

from itertools import takewhile

class Foo:
    def __init__(self, K = None):
        self.bf=open('b', 'w')
        self.cf=open('c', 'w')
        self.count = 0
        self.K = K

    def Go(self):
        for self.line in takewhile(self.Lamda(), open('a')):
            self.SplitLine()
            if self.IsValidPid():
                self.WriteLineToFiles()

    def SplitLine(self):
        self.lineSplit=self.line.split(',')

    def Lamda(self):
        if self.K is None:
            return lambda x: x.split(',')[0] != '[Controls]'
        else:
            return lambda x: x.split(',')[0] != '[Controls]' and self.count < self.K

    def IsValidPid(self):
        pid=self.lineSplit[1][0:3]
        return pid!='cnv' and pid!='hCV' and pid!='cnv'

    def WriteLineToFiles(self):
        self.count += 1
        self.bf.write(self.ParseLine())
        if self.K is None:
            self.cf.write(self.line)

    def ParseLine(self):
        return (self.lineSplit[1] + ',' + self.lineSplit[2] + ',' + 
                self.lineSplit[3][1] + self.lineSplit[3][3] + ',' +
                self.lineSplit[15] + ',' + self.lineSplit[9] + ',' + 
                self.lineSplit[10]).strip('"')+'\n'

Foo().Go()

原始版本：

from itertools import takewhile

if K is None:
    illuminacond = lambda x: x.split(',')[0] != '[Controls]'
else:
    illuminacond = lambda x: x.split(',')[0] != '[Controls]' and i < K

def Parse(line):
    return (line[1] + ',' + line[2] + ',' + line[3][1] + line[3][3] + ',' +
            line[15] + ',' + line[9] + ',' + line[10]).strip('"')+'\n'

def IsValidPid(line_split):
    pid=line_split[1][0:3]
    return pid!='cnv' and pid!='hCV' and pid!='cnv'

bf=open('b', 'w')
cf=open('c', 'w')

def WriteLineToFiles(line, line_split):
    bf.write(Parse(line_split))
    if K is None:
        cf.write(line)

i = 0

for line in takewhile(illuminacond, open('a')):
    line_split=line.split(',')
    if IsValidPid(line_split):
        WriteLineToFiles(line, line_split)
        i += 1

避免Python代码redux中的代码重复

4 个答案: