Question

我有两个清单：

a_list包含不同的tupled id：

a_list = [('1'), ('2'), ('3'), ('4')...etc]

并列在名为check_list的元组列表中：

check_list = [[[('1'), ('2')]], [[('2'), ('3')], [('3'),('4')]], [[('1'),('3')]]...etc]

我的问题对于我的python技能级别来说有点过于复杂......我正在尝试形成一个矩阵输出到csv文件，具有以下结构：

     1 2 3 4

1 1 1 0 0

2 0 1 2 1

3 1 0 1 0

其中每个值都是a_list中check_list的数字的计数。

我看过numpy而且我不知道。我和this一起玩过，但我似乎无法完全理解将其转移到我的问题上。我对阵列知识也非常有限。提前谢谢。

Answer 1

您可以使用Counter dict获取计数，按键排序以使用OrderedDict获取Order中的计数以维护该顺序并使用document.querySelectorAll("td[id^=cell]")查找缺少的键。

dict.viewkeys

输出：

from __future__ import print_function
from collections import Counter, OrderedDict
from itertools import chain

check_list = [[[('1',), ('2',)]], [[('2',), ('3',)], [('3',), ('4',)]], [[('1',), ('3',)]]]

a_list = [('1',), ('2',), ('3',), ('4',)]

cn = OrderedDict(sorted(Counter(a_list).items()))

print(" ".join([str(t[0]) for t in cn]))
for chk in check_list:
    _cn = Counter(chain(*chk))
    # cn.keys() python 3
    diff = cn.viewkeys() - _cn
    for k  in cn:
        if k not in diff:
            print(_cn[k], end=" ")
        else:
            print(0, end=" ")
    print()

如果您不关心订单，可以删除sorted / Ordereddict逻辑：

这会给你一个任意的顺序：

from collections import Counter
from itertools import chain

check_list = [[[('1',), ('2',)]], [[('2',), ('3',)], [('3',), ('4',)]], [[('1',), ('3',)]]]

a_list = [('1',), ('2',), ('3',), ('4',)]

cn = Counter(a_list)

print(" ".join([str(t[0]) for t in cn]))
for chk in check_list:
    _cn = Counter(chain(*chk))
    diff = cn.viewkeys() - _cn
    for k  in cn:
        if k not in diff:
            print(_cn[k], end=" ")
        else:
            print(0, end=" ")
    print()

写入csv：

out.csv：

from collections import Counter, OrderedDict
from itertools import chain
from csv import writer

check_list = [[[('1',), ('2',)]], [[('2',), ('3',)], [('3',), ('4',)]], [[('1',), ('3',)]]]

a_list = [('1',), ('2',), ('3',), ('4',)]

with open("out.csv","w") as out:
    wr = writer(out,delimiter=" ")
    cn = OrderedDict(sorted(Counter(a_list).items()))
    wr.writerow(list(chain(*cn)))
    for chk in check_list:
        _cn = Counter(chain(*chk))
        diff = cn.viewkeys() - _cn
        wr.writerow([_cn[k] if k not in diff else 0 for k in cn])

Python 2.7从列表创建矩阵

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