Python：统计表格的所有可能组合

Question

Python：统计表格的所有可能组合

8

我有一个表格，它长这样：

   PotA  PotB  PotC  PotD  PotE
A   +     +     +     +     +
B   -     ?     +     +     ?
C   +     +     +     +     +
D   +     -     +     -     +
E   +     +     +     +     +

从这里开始，我需要找到所有可能的“+”、“-”和“？”组合，针对(PotA和PotB)、(PotA和PotC)等所有组合，一直到(PotA、PotB、PotC、PotD和PotE)。实际上，“Pot”行还可以继续下去，但为了简化起见，我只展示到PotE。

为了做到这一点，首先，我按照以下方式读取文件，然后生成两个组合的所有可能性，以计算每个可能性的数量。

def readDatafile():
    filename = ("data.txt")
    infile = open(filename,'r')

    for line in infile.readlines():
        line = line.strip()
        print (line)          # just to check the data, need to carry on from here.

"""Generate all possible permutations for later count"""
def doPermutations(items, n):
    if n == 0:
        yield ''
    else:
        for i in range(len(items)):
            for base in doPermutations(items, n - 1):
                yield str(items[i]) + str(base)

def makeAllPossibleList():
    nlength      = 2          # This should go outside of the function and will be same as the number of Pots
    lpossibility = ['+', '-', '?']
    litems       = []

    for i in doPermutations(lpossibility, int(nlength)):
        litems.append(i)

    for x in items:
        print (x)             # This generate all the possible items for combination of two

所以，最终结果将会是这样的：

Combination: Possibility Count
PotA, PotB: ++ 3
PotA, PotB: +- 1
PotA, PotB: +? 0
PotA, PotB: -+ 0
PotA, PotB: -- 0
PotA, PotB: -? 1
PotA, PotB: ?+ 0
PotA, PotB: ?- 0
PotA, PotB: ?? 0
PotA, PotC: ...
PotA, PotC: ...
.......
PotA, PotB, PotC, PotD, PotE: +++++ 3
PotA, PotB, PotC, PotD, PotE: ++++- 0
PotA, PotB, PotC, PotD, PotE: ++++? 0
.......

有没有好的Python方法来解决这个问题？我是否需要将带有标题的数据读取为键和列值的列表，才能得到正确的逻辑？

我无法得到正确的逻辑，请给我一些帮助。

- Karyo

6

请访问http://docs.python.org/3.3/library/itertools.html，特别是http://docs.python.org/3.3/library/itertools.html#itertools.permutations - user1907906

好的，我正在努力学习。 - Karyo

1

collections.Counter 或 collections.defaultdict 可以帮助进行计数。 - user2357112

1个回答

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- DSM · Accepted Answer

假设我理解你的意思，那么可以考虑这样做：

import itertools
import collections

def read_table(filename):
    with open(filename) as fp:
        header = next(fp).split()
        rows = [line.split()[1:] for line in fp if line.strip()]
        columns = zip(*rows)
    data = dict(zip(header, columns))
    return data

table = read_table("data.txt")
pots = sorted(table)

alphabet = "+-?"
for num in range(2, len(table)+1):
    for group in itertools.combinations(pots, num):
        patterns = zip(*[table[p] for p in group])
        counts = collections.Counter(patterns)
        for poss in itertools.product(alphabet, repeat=num):
            print ', '.join(group) + ':',
            print ''.join(poss), counts[poss]

它会产生：

PotA, PotB: ++ 3
PotA, PotB: +- 1
PotA, PotB: +? 0
PotA, PotB: -+ 0
PotA, PotB: -- 0
PotA, PotB: -? 1
PotA, PotB: ?+ 0
PotA, PotB: ?- 0
PotA, PotB: ?? 0
PotA, PotC: ++ 4
[...]
PotA, PotB, PotC, PotD, PotE: +++++ 3
PotA, PotB, PotC, PotD, PotE: ++++- 0
[...]

请注意，我假设你的期望输出是错误的，因为在这行代码中：

PotA, PotB, PotC, PotD, PotE: ++++++ 2

左边有五列，但右边有六个加号符号。