如何使新的数字列表不附加到自己的列表中？

Question

如果我有一个称为t的多维列表，并且我将列表中的一些数字附加到一个称为TC的新列表中，那么如何将所有没有附加到新列表中的数字放入它们自己的列表中，称为nonTC？例如：

t = [[1, 3, 4, 5, 6, 7],[9, 7, 4, 5, 2], [3, 4, 5]]

我写了一些条件，只从每个列表中添加一些值来创建新的列表TC：

TC = [[3, 4, 6], [9, 7, 2], [5]]

如何将TC中未包含的值附加到它自己的列表中？所以我会得到：

nonTC = [[1, 5, 7],[4, 5],[3,4]]

Answer 1

只是出于好奇，使用NumPy：

import numpy as np

t = [[1, 3, 4, 5, 6, 7],[9, 7, 4, 5, 2], [3, 4, 5]]
TC = [[3, 4, 6], [9, 7, 2], [5]]

print([np.setdiff1d(a, b) for a, b in zip(t, TC)])
#=> [array([1, 5, 7]), array([4, 5]), array([3, 4])]

Answer 2

您可以使用列表理解和集合列表来筛选原始列表：

t = [[1, 3, 4, 5, 6, 7],[9, 7, 4, 5, 2], [3, 4, 5]]

# filter sets - each index corresponds to one inner list of t - the numbers in the 
# set should be put into TC - those that are not go into nonTC
getem = [{3,4,6},{9,7,2},{5}]

TC = [ [p for p in part if p in getem[i]] for i,part in enumerate(t)]
print(TC)

nonTC = [ [p for p in part if p not in getem[i]] for i,part in enumerate(t)]     
print(nonTC)

输出：

[[3, 4, 6], [9, 7, 2], [5]] # TC
[[1, 5, 7], [4, 5], [3, 4]] # nonTC

读数：

• list comprehensions
• sets
• enumerate(iterable)

和：Explanation of how nested list comprehension works?

---

其他方法的建议，AChampion的信条：

TC_1 = [[p for p in part if p in g] for g, part in zip(getem, t)]
nonTC_1 = [[p for p in part if p not in g] for g, part in zip(getem, t)]

请参见zip()——它本质上是将两个列表捆绑成一个元组的可数

( (t[0],getem[0]), (t[1],getem[1]) (t[2],getem[2])) 

---

多个事件的附加组件-没收列表组件和集合：

t = [[1, 3, 4, 5, 6, 7, 3, 3, 3],[9, 7, 4, 5, 2], [3, 4, 5]]

# filter lists - each index corresponds to one inner list of t - the numbers in the list
# should be put into TC - those that are not go into nonTC - exactly with the amounts given
getem = [[3,3,4,6],[9,7,2],[5]]

from collections import Counter
TC = []
nonTC = []
for f, part in zip(getem,t):
    TC.append([])
    nonTC.append([])
    c = Counter(f) 
    for num in part:
        if c.get(num,0) > 0:
            TC[-1].append(num)
            c[num]-=1
        else:
            nonTC[-1].append(num)            

print(TC)    # [[3, 4, 6, 3], [9, 7, 2], [5]]
print(nonTC) # [[1, 5, 7, 3, 3], [4, 5], [3, 4]]

它只需要一个通过你的项目，而不是2（单独的列表comps），这使它可能更有效地从长远来看