如何合并两个列表？保留相同的列表元素以进行集合操作

jmunsch 发表于 Dev

蒙施

我一直在画维恩图，编码循环和不同的集合（symmetrical_differences，并集，交集，isdisjoint），在一两天的大部分时间里用行号枚举，试图找出如何在代码中实现这一点。

a = [1, 2, 2, 3] # <-------------|
b = [1, 2, 3, 3, 4] # <----------| Do not need to be in order.
result = [1, 2, 2, 3, 3, 4] # <--|

要么：

A = [1,'d','d',3,'x','y']
B = [1,'d',3,3,'z']
result =  [1,'d','d',3,3,'x','y','z']

编辑：

不尝试做a + b= [1、2、2、2、2、3、3、3、4]

尝试做类似的事情：

a - b = [2]

b - a = [3，4]

a ∩ b = [1,2,3]

所以

[a - b] + [b - a] + a ∩ b = [1,2,2,3,3,4]？

我不确定在这里。

我有两个电子表格，每个电子表格都有几千行。我想按列类型比较两个电子表格。

我已经从每一列中创建了要比较/合并的列表。

def returnLineList(fn):
    with open(fn,'r') as f:
        lines = f.readlines()
    line_list = []
    for line in lines:
        line = line.split('\t')
        line_list.append(line)
    return line_list

def returnHeaderIndexDictionary(titles):
    tmp_dict = {}
    for x in titles:
        tmp_dict.update({x:titles.index(x)})
    return tmp_dict

def returnColumn(index, l):
    column = []
    for row in l:
        column.append(row[index])
    return column

def enumList(column):
    tmp_list = []
    for row, item in enumerate(column):
        tmp_list.append([row,item])
    return tmp_list

def compareAndMergeEnumerated(L1,L2):
    less = []
    more = []
    same = []
    for row1,item1 in enumerate(L1):
        for row2,item2 in enumerate(L2):
            if item1 in item2:
                count1 = L1.count(item1)
                count2 = L2.count(item2)
                dif = count1 - count2
                if dif != 0:
                    if dif < 0:
                        less.append(["dif:"+str(dif),[item1,row1],[item2,row2]])
                    if dif > 0:
                        more.append(["dif:"+str(dif),[item1,row1],[item2,row2]])
                else:
                    same.append(["dif:"+str(dif),[item1,row1],[item2,row2]])
                break
    return less,more,same,len(less+more+same),len(L1),len(L2)

def main():
    unsorted_lines = returnLineList('unsorted.csv')
    manifested_lines = returnLineList('manifested.csv')

    indexU = returnHeaderIndexDictionary(unsorted_lines[0])
    indexM = returnHeaderIndexDictionary(manifested_lines[0])

    u_j_column = returnColumn(indexU['jnumber'],unsorted_lines)
    m_j_column = returnColumn(indexM['jnumber'],manifested_lines)

    print(compareAndMergeEnumerated(u_j_column,m_j_column))

if __name__ == '__main__':
    main()

更新：

from collections import OrderedDict
A = [1,'d','d',3,'x','y']
B = [1,'d',3,3,'z']
M = A + B
R = [1,'d','d',3,3,'x','y','z']


ACount = {}
AL = lambda x: ACount.update({str(x):A.count(x)})
[AL(x) for x in A]

BCount = {}
BL = lambda x: BCount.update({str(x):B.count(x)})
[BL(x) for x in B]

MCount = {}
ML = lambda x: MCount.update({str(x):M.count(x)})
[ML(x) for x in M]


RCount = {}
RL = lambda x: RCount.update({str(x):R.count(x)})
[RL(x) for x in R]


print('^sym_difAB',set(A) ^ set(B)) # set(A).symmetric_difference(set(B))
print('^sym_difBA',set(B) ^ set(A)) # set(A).symmetric_difference(set(B))
print('|union    ',set(A) | set(B)) # set(A).union(set(B))
print('&intersect',set(A) & set(B)) # set(A).intersection(set(B))
print('-dif AB   ',set(A) - set(B)) # set(A).difference(set(B))
print('-dif BA   ',set(B) - set(A)) 
print('<=subsetAB',set(A) <= set(B)) # set(A).issubset(set(B))
print('<=subsetBA',set(B) <= set(A)) # set(B).issubset(set(A))
print('>=supsetAB',set(A) >= set(B)) # set(A).issuperset(set(B))
print('>=supsetBA',set(B) >= set(A)) # set(B).issuperset(set(A))

print(sorted(A + [x for x in (set(A) ^ set(B))]))
#[1, 3, 'd', 'd', 'x', 'x', 'y', 'y', 'z']

print(sorted(B + [x for x in (set(A) ^ set(B))]))
#[1, 3, 3, 'd', 'x', 'y', 'z', 'z']
cA = lambda y: A.count(y)
cB = lambda y: B.count(y)
cM = lambda y: M.count(y)
cR = lambda y: R.count(y)
print(sorted([[y,cA(y)] for y in (set(A) ^ set(B))]))
#[['x', 1], ['y', 1], ['z', 0]]

print(sorted([[y,cB(y)] for y in (set(A) ^ set(B))]))
#[['x', 0], ['y', 0], ['z', 1]]

print(sorted([[y,cA(y)] for y in A]))
print(sorted([[y,cB(y)] for y in B]))
print(sorted([[y,cM(y)] for y in M]))
print(sorted([[y,cR(y)] for y in R]))
#[[1, 1], [3, 1], ['d', 2], ['d', 2], ['x', 1], ['y', 1]]
#[[1, 1], [3, 2], [3, 2], ['d', 1], ['z', 1]]
#[[1, 2], [1, 2], [3, 3], [3, 3], [3, 3], ['d', 3], ['d', 3], ['d', 3], ['x', 1], ['y', 1], ['z', 1]]
#[[1, 1], [3, 2], [3, 2], ['d', 2], ['d', 2], ['x', 1], ['y', 1], ['z', 1]]

cAL = sorted([[y,cA(y)] for y in A])

在此处输入图片说明

更新：2

基本上我认为是时候该学习了：

大熊猫

它看起来像是聚合，分组和求和的组合。

亚伦·霍尔

经过进一步的审查（现在我已经在家中并尝试使用Python解释器），我了解到您正在尝试执行的操作，但这与删除重复项的标题相矛盾。我看到您正在将每个其他元素视为一个新的索引唯一项。

这在概念上类似于装饰，排序，未装饰模式，只是用“连接”或“设置操作”代替术语“排序”。

因此，这是一个设置，首先导入，itertools因此我们可以将每个相似的元素分组并将它们枚举为一组：

import itertools

def indexed_set(a_list):
    '''
    assuming given a sorted list, 
    groupby like items, 
    and index from 0 for each group
    return a set of tuples with like items and their index for set operations
    '''
    return set((like, like_index) for _like, like_iter in itertools.groupby(a_list)
                          for like_index, like in enumerate(like_iter))

稍后，我们需要将带有索引的集合转换回列表：

def remove_index_return_list(an_indexed_set):
    '''
    given a set of two-length tuples (or other iterables)
    drop the index and 
    return a sorted list of the items 
    (sorted by str() for comparison of mixed types)
    '''
    return sorted((item for item, _like_index in an_indexed_set), key=str)

最后，我们需要我们的数据（取自您提供的数据）：

a = [1, 2, 2, 3] 
b = [1, 2, 3, 3, 4] 
expected_result = [1, 2, 2, 3, 3, 4]

这是我建议的用法：

a_indexed = indexed_set(a)
b_indexed = indexed_set(b)
actual_result = remove_index_return_list(a_indexed | b_indexed)

assert expected_result == actual_result

不会引发AssertionError，并且

print(actual_result)

印刷品：

[1, 2, 2, 3, 3, 4]

编辑：由于我使函数处理混合大小写，所以我想演示：

c = [1,'d','d',3,'x','y']
d = [1,'d',3,3,'z']
expected_result =  [1,'d','d',3,3,'x','y','z']
c_indexed = indexed_set(c)
d_indexed = indexed_set(d)
actual_result = remove_index_return_list(c_indexed | d_indexed)
assert actual_result == expected_result

而且我们看到的并没有我们所期望的完全一样，但是由于排序的缘故，结果非常接近：

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AssertionError
>>> actual_result
[1, 3, 3, 'd', 'd', 'x', 'y', 'z']
>>> expected_result
[1, 'd', 'd', 3, 3, 'x', 'y', 'z']

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