如何根据元素值组合列表中的元素?

Question

如果我想根据元素值在列表内组合列表，我该如何实现?假设if list

lis = [['steve','reporter','12','34','22','98'],['megan','arch','44','98','32','22'],['jack','doctor','80','32','65','20'],['steve','dancer','66','31','54','12']]

这里包含'steve'的列表出现了两次，所以我想将它们组合在一起

new_lis = [['steve','reporter','12','34','22','98','dancer','66','31','54','12'],['megan','arch','44','98','32','22'],['jack','doctor','80','32','65','20']]

我尝试下面的代码来实现这个

new_dic = {}
for i in range(len(lis)):
name = lis[i][0]
if name in new_dic:
new_dic[name].append([lis[i][1],lis[i][2],lis[i][3],lis[i][4],lis[i][5]])
else:
new_dic[name] = [lis[i][1],lis[i][2],lis[i][3],lis[i][4],lis[i][5]]
print(new_dic)

我最终创建了一个包含多个列表值的字典，如下所示

{'steve': ['reporter', '12', '34', '22', '98', ['dancer', '66', '31', '54', '12']], 'megan': ['arch', '44', '98', '32', '22'], 'jack': ['doctor', '80', '32', '65', '20']}

但是我想把它作为一个列表，这样我就可以转换成下面的格式

new_lis = [['steve','reporter','12','34','22','98','dancer','66','31','54','12'],['megan','arch','44','98','32','22'],['jack','doctor','80','32','65','20']]

是否有一种不同的方法来解决这个问题?

Answer 1

使用groupby函数和itertools函数有不同的方法。还有一些方法可以将字典转换为列表。这完全取决于你想要什么。

from itertools import groupby
lis = [['steve','reporter','12','34','22','98'],['megan','arch','44','98','32','22'],['jack','doctor','80','32','65','20'],['steve','dancer','66','31','54','12']]
lis.sort(key = lambda x: x[0])
output = []
for name , groups in groupby(lis, key = lambda x: x[0]):
temp_list = [name]
for group in groups:
temp_list.extend(group[1:])
output.append(temp_list)
print(output)

输出

[['jack', 'doctor', '80', '32', '65', '20'], ['megan', 'arch', '44', '98', '32', '22'], ['steve', 'reporter', '12', '34', '22', '98', 'dancer', '66', '31', '54', '12']] 

Answer 2

不确定这个片段是否回答了您的问题。就时间复杂度而言，这不是最快的方法。如果我有更好的解决方法，我会更新这个答案。

lis = [['steve','reporter','12','34','22','98'],['megan','arch','44','98','32','22'],['jack','doctor','80','32','65','20'],['steve','dancer','66','31','54','12']]
new_lis = []
element_value = 'steve'
for inner_lis in lis:
if element_value in inner_lis:
if not new_lis: 
new_lis+=inner_lis
else:
inner_lis.remove(element_value)
new_lis+=inner_lis
lis.remove(inner_lis)
print([new_lis] + lis)

输出

[['steve', 'reporter', '12', '34', '22', '98', 'dancer', '66', '31', '54', '12'], ['megan', 'arch', '44', '98', '32', '22'], ['jack', 'doctor', '80', '32', '65', '20']]