从列表创建一个子列表

Question

我有一个Training_list，它的列表列表例如

[[1,2,3,4,5,6,7,8,9,10,11,12,13,'<50k'],[1,2,3,4,5,6,7,8,9,10,11,12,13,'<50k'], 
[1,2,3,4,5,6,7,8,9,10,11,12,13,'>50k'],[1,2,3,4,5,6,7,8,9,10,11,12,13,'>50k'], 
[1,2,3,4,5,6,7,8,9,10,11,12,13,'<50k'],[1,2,3,4,5,6,7,8,9,10,11,12,13,'>50k'], 
... 
] 

我想根据最后一个属性将这个列表分成两个子列表。 第一个列表应包含所有的< 50k记录的under_50k列表清单，例如，

[[1,2,3,4,5,6,7,8,9,10,11,12,13], [1,2,3,4,5,6,7,8,9,10,11,12,13], ...] 

第二个列表应包含所有> 50k记录的over_50k列表列表，例如，

[[1,2,3,4,5,6,7,8,9,10,11,12,13], [1,2,3,4,5,6,7,8,9,10,11,12,13], ...] 

一旦两个列表创建然后我试图加起来每个索引列表 例如

[1,2,3,4,5,6,7,8,9,10,11,12,13] + [1,2,3,4,5,6,7,8,9,10,11,12,13] 
= [2,4,6,8,10,12,14,16,18,20,22,24,26] 

似乎可以得到清单的细分工作。

def sums_list(): 

    sums_list = [] 
    try: 
     for index in range(15): 
      sums_list.append(under_50k_list[index]+over_50k_list[index]) 
    except: 
     pass 
     return(sums_list) 

def under_over_lists(): 

    under_50k_list = [0]*14 
    under_50k_count = 0 
    over_50k_list = [0]*14 
    over_50k_count = 0 
    try: 
     for row in training_list: 
      if row[-1].lstrip() == '<=50K': 
       under_50k_list = sums_list(under_50k_list, row[:-1]) 
       under_50k_count += 1 
      else: 
       if row[-1].lstrip() == '>50K': 
        over_50k_list = sums_list(over_50k_list, row[:-1]) 
        over_50k_count += 1 
    except: 
     pass 
     print(under_50k_list) 
     return under_over_lists 

Answer 1

您可以使用numpy如果list的每个子列表是同样大小的

>>> import numpy as np 
>>> llist=[[1,2,3,4,5,6,7,8,9,10,11,12,13,'<50k'],[1,2,3,4,5,6,7,8,9,10,11,12,13,'<50k'], [1,2,3,4,5,6,7,8,9,10,11,12,13,'>50k'],[1,2,3,4,5,6,7,8,9,10,11,12,13,'>50k'],[1,2,3,4,5,6,7,8,9,10,11,12,13,'<50k'],[1,2,3,4,5,6,7,8,9,10,11,12,13,'>50k']] 
>>> under_50k_list=[i[:-1] for i in llist if i[-1]=='<50k'] 
>>> under_50k_list 
[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]] 
>>> over_50k_list=[i[:-1] for i in llist if i[-1]=='>50k'] 
>>> over_50k_list 
[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]] 
>>> sum(np.array(under_50k_list)) 
array([ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39]) 
>>> under_50k_sum=sum(np.array(under_50k_list)) 
>>> over_50k_sum=sum(np.array(over_50k_list)) 
>>> under_50k_sum 
array([ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39]) 
>>> over_50k_sum 
array([ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39]) 

Answer 2

您应该使用append方法，让您的细分正常工作。我觉得它对动态大小的列表更友好。

over_50k = [] 
under_50k = [] 
for row in training_list: 
    if row[-1] == "<50k": 
     under_50k.append(row[:-1]) 
    elif row[-1] == ">50k": 
     over_50k.append(row[:-1]) 

现在，让您的资金：

over_50k_sum = [i for i in over_50k[0]] # initialize with the first one 
for i in range(1,len(over_50k)):   # skips the first one 
    for j in range(len(over_50k[i])): 
     over_50k_sum[j] += over_50k[i][j] 

under_50k_sum = [i for i in under_50k[0]] # initialize with the first one 
for i in range(1,len(under_50k)):   # skips the first one 
    for j in range(len(under_50k[i])): 
     under_50k_sum[j] += under_50k[i][j] 

Answer 3

假设你只想要的结果，而不是中间的列表，它只是：

ll = [[1,2,3,4,5,6,7,8,9,10,11,12,13,'<50k'], ... ] 

sumlist = lambda a,b:[x+y for x,y in zip(a,b)] 
def sum_if(lists, key): 
    return reduce(sumlist, (l[:-1] for l in lists if l[-1]==key)) 

under_50k_count = sum_if(ll, '<50k') 
over_50k_count = sum_if(ll, '>50k') 

这可能是值得导入ìzip from itertools和如果您的列表很长并且想要减少复制，那么使用该代替zip，但它肯定不是必需的。

Answer 4

由于单行的乐趣：

ll = [[1,2,3,4,5,6,7,8,9,10,11,12,13,'<50k'], ...] 
under_50k_count = [sum(b) for b in zip(*[a[:-1] for a in ll if a[-1].startswith('<')])] 
over_50k_count = [sum(b) for b in zip(*[a[:-1] for a in ll if a[-1].startswith('>')])] 

这本身是不是一个有用的堆栈溢出的答案，落入类别“试试这个”没有解释，让我们打破它一点。

我们用列表解析分离出名单的2个不同的种类：

[a[:-1] for a in ll if a[-1].startswith('<')] 

我们再unpack这个名单并把它传递给压缩，这为我们提供了一个元组的列表：

[(1,1,1,...), (2,2,2,...), ...] 

然后，我们使用另一个列表理解来对这些元组进行求和。

列表解析明显很快，解包列表不是（尤其是大列表）。所以，虽然压缩这样的东西很有趣，但如果速度有任何问题，或者如果任何人继承你的代码甚至有远程可能性，我都不会推荐使用它。