Question

例如，如果我的嵌套列表是：

[['2HC'], ['4BB'], ['4BB'], ['2HC']]

如何将['2HC']引用到'A'，将['4BB']引用到'B'，这样我的结果将是：

['A', 'B', 'B', 'A']

Answer 1

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       <option>1</option>
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     <select id="select2">
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输出：

l = [['2HC',], ['4BB'], ['4BB'], ['2HC']]
mapping = {'2HC' : 'A', '4BB': 'B'}

new_list = [mapping[i[0]] for i in l]
print(new_list)

Answer 2

l = [['2HC'], ['4BB'], ['4BB'], ['2HC']]
d =  {'2HC' : 'A', '4BB': 'B'}

map(lambda x: d[x[0]], l)
# ['A', 'B', 'B', 'A']

Answer 3

l = [['2HC'], ['4BB'], ['4BB'], ['2HC']]

lst = ['A' if x[0] == '2HC' else 'B' for x in l]

print(lst)

输出

['A', 'B', 'B', 'A']

Answer 4

您可以拉平列表以使其更容易，然后使用简单的翻译映射：

from itertools import chain

nested_vs = [['2HC'], ['4BB'], ['4BB'], ['2HC']]

# flatten
vs = chain.from_iterable(nested_vs)

# translate
translations = {'2HC' : 'A', '4BB': 'B'}
translated = list(map(translations.get, vs))

现在translated保持：

['A', 'B', 'B', 'A']

Answer 5

针对此问题的itertools解决方案-

data = [['2HC',], ['4BB'], ['4BB'], ['2HC']]
mapping = {'2HC' : 'A', '4BB': 'B'}
l = [mapping[i] for i in itertools.chain.from_iterable(data)]

itertools的好处是，即使内部列表包含多个元素，它也可以将列表数据转换为平面列表。

例如，列表data是-

[['2HC', '2HC'], ['4BB'], ['4BB'], ['2HC']]

我们仍然会获得正确的输出。

如何将嵌套列表的列表更改为字符串？

5 个答案: