是否可以保存和访问包含不同长度元素的数组或列表?例如,我想保存data=[s,r,a,se]
r,a是标量但是s
和se
是一个包含4个元素的数组。(用python语言)
例如,在一次中:(s,r,a,se)
在不同时间有所不同
s=[1,3,4,6] r=5 a=7 se=[11,12,13,14]
data=[s,r,a,se]=[[1,3,4,6],5,7,[11,12,13,14]]
如何定义包含它们的数组,以便能够调用它们,类似于以下代码:
s=[data[0] for data in minibatch]
r=[data[1] for data in minibatch]
a=[data[2] for data in minibatch]
se=[data[3] for data in minibatch]
此外,我如何提取(查找)[stest,rtest,atest,setest]
中的特殊data
(stest,setest
有4个元素)
例如:我希望看到我能在数据中找到[[1,2,3,4],5,6,[7,8,9,10]]
,类似于:[ [[1,2,3,4],5,6,[7,8,9,10]] ,[[...,...,...,...],...,..., [...,...,...,...]], [[18,20,31,42],53,666,[27,48,91,120]]]
如果我没有找到,我会附加它,否则什么也没发生!
答案 0 :(得分:1)
您可以将它们添加到新列表中:
new_list = [s, r, a, se]
但是你必须小心管理这个清单
答案 1 :(得分:0)
# This is a great opportunity to explore dictionaries
# lets take the examples you have given in variales
s=[1,3,4,6]
r=5
a=7
se=[11,12,13,14]
# make a dictionary out of them, with keys which are
# the same as the variable name
my_dict = {'s':s,
'r':r,
'a':a,
'se':se}
# lets see what that looks like
print(my_dict)
print()
# to retrieve 2nd (=ix=1) element of s
# the format to do this is simple
# ditionary_variable_name['the string which is the key'][index_of_the_list]
s2 = my_dict['s'][1]
print(s2)
print()
# to retrieve all of se
se_retrieved = my_dict['se']
print(se_retrieved)
# you can further experiment with this
样本输出:
{'s': [1, 3, 4, 6], 'r': 5, 'a': 7, 'se': [11, 12, 13, 14]}
3
[11, 12, 13, 14]
为了写这篇文章,你需要做这样的事情:
my_dict['se'] = [15, 16, 17, 18]
或
my_dict['se'][2] = 19