python - 代码审查 : Python list concatenation elements to string

Question

嘿，我有一个问题，正在解决一个问题看起来像这样:

输出正是我所需要的，但我想知道是否没有更优雅的如何做到这一点？

list = [["key189","0","apples"],
        ["key622","1","bananas"],
        ["key233","2","bacon"],
        ["key454","3","bread"],
        ["key35","4","jam"],
        ["key6","5","coffee"]]

for e in list:
    if e[0] == "key622":
        key622 = e[2]
    if e[0] == "key189":
        key189 = e[2]
    if e[0] == "key35":
        key35 = e[2]
    if e[0] == "key454":
        key454 = e[2]
    if e[0] == "key233":
        key233 = e[2]
    if e[0] == "key6":
            key6 = e[2]

string_form = "|".join([key6, key35, key233, key189, key622, key454])


print(string_form)
# prints coffee|jam|bacon|apples|bananas|bread

Answer 1

我认为使用字典会让你的生活更轻松。

dt = {itm[0]:itm[2] for itm in list}

dt 将是:

{'key189': 'apples',
 'key233': 'bacon',
 'key35': 'jam',
 'key454': 'bread',
 'key6': 'coffee',
 'key622': 'bananas'}

string_form = "|".join(dt.values())

string_form 将是:

'coffee|apples|bacon|jam|bananas|bread'

你可以获得任何键的值:像这样:

dt.get('key35')

对于特定字符串，请使用:

"|".join([dt.get('key6'), dt.get('key35'), dt.get('key233'), dt.get('key189'), dt.get('key622'), dt.get('key454')])

尽量不要使用list、dict等变量名