python - 根据映射从给定列表生成列表的 Python 程序

Question

例如

组织列表：

aa b2 c d

映射：

aa 1
b2 2
d 3
c 4

gen_list：

1 2 4 3

Python 的实现方式是什么？假设 org_list 和映射在 filesorg_list.txt和mapping.txt中，而 gen_list 将被写入gen_list.txt

顺便说一句，您希望哪种语言很容易实现这一点？

Answer 1

只需使用列表推导遍历列表：

gen_list = [mapping[i] for i in org_list]

演示：

>>> org_list = ['aa', 'b2', 'c', 'd']
>>> mapping = {'aa': 1, 'b2': 2, 'd': 3, 'c': 4}
>>> [mapping[i] for i in org_list]
[1, 2, 4, 3]

如果您在文件中有此数据，请首先在内存中构建映射：

with open('mapping.txt') as mapfile:
    mapping = {}
    for line in mapfile:
        if line.strip():
            key, value = line.split(None, 1)
            mapping[key] = value

然后从输入文件构建输出文件：

with open('org_list.txt') as inputfile, open('gen_list.txt', 'w') as outputfile:
    for line in inputfile:
        try:
            outputfile.write(mapping[line.strip()] + '\n')
        except KeyError:
            pass  # entry not in the mapping

Answer 2

这是您的情况的解决方案。

with open('org_list.txt', 'rt') as inp:
    lines = inp.read().split()
    org_list = map(int, lines)

with open('mapping.txt', 'rt') as inp:
    lines = inp.readlines()
    mapping = dict(line.split() for line in lines)

gen_list = (mapping[i] for i in org_list) # Or you may use `gen_list = map(mapping.get, org_list)` as suggested in another answers

with open('gen_list.txt', 'wt') as out:
    out.write(' '.join(gen_list))

我认为 Python 足够优雅地处理这种情况。

Answer 3

另一种方式：

In [1]: start = [1,2,3]
In [2]: mapping = {1: "one", 2: "two", 3: "three"}
In [3]: map(mapping.get, start)
Out[3]: ['one', 'two', 'three']

Answer 4

尝试使用map()或列表推导：

>>> org_list = ['aa', 'b2', 'c', 'd']
>>> mapping = {'aa': 1, 'b2': 2, 'd': 3, 'c': 4}

>>> map(mapping.__getitem__, org_list)
[1, 2, 4, 3]

>>> [mapping[x] for x in org_list]
[1, 2, 4, 3]

Answer 5

mapping = dict(zip(org_list, range(1, 5)))       # change range(1, 5) to whatever
gen_list = [mapping[elem] for elem in org_list]  # you want it to be