英文:
Add a column of string using list's indices from another column
问题
idx score name0 (3,) 0.773 (D,)1 (3, 5) 0.841 (D, F)2 (1, 3, 5) 0.862 (B, D, F)3 (1, 3, 5, 10) 0.874 (B, D, F, K)4 (1, 3, 5, 8, 10) 0.883 (B, D, F, I, K)
英文:
Having this list of name:
name_list = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K']
As well as the following df:
df = pd.DataFrame({'idx': ['(3,)','(3, 5)','(1, 3, 5)','(1, 3, 5, 10)','(1, 3, 5, 8, 10)'],'score': [0.773,0.841,0.862,0.874,0.883]})df.head(2)idx score0 (3,) 0.7731 (3, 5) 0.841
The idx column represents indices of the elements of name_list. I want to add a new column name to the df with the corresponding name from the list.
Expected results:
idx score name0 (3,) 0.773 (D,)1 (3, 5) 0.841 (D, F)2 (1, 3, 5) 0.862 (B, D, F)3 (1, 3, 5, 10) 0.874 (B, D, F, K)4 (1, 3, 5, 8, 10) 0.883 (B, D, F, I, K)
答案1
得分: 1
这是你要的翻译结果：
需要执行几个步骤：- 创建一个用于将索引映射到列表值的映射字典，- 使用 [`ast.literal_eval`](https://docs.python.org/3/library/ast.html#ast.literal_eval) 将元组的字符串表示转换为元组，- 使用元组推导式将值进行映射```pythonfrom ast import literal_evald = dict(enumerate(name_list))df['name'] = [tuple(d.get(x, '?') for x in literal_eval(t))for t in df['idx']]
如果您确信索引是有效的，无需使用字典：
df['name'] = [tuple(name_list[x] for x in literal_eval(t))for t in df['idx']]
如果需要字符串输出：
df['name'] = [f"({' '.join(tuple(name_list[x] for x in literal_eval(t)))})"for t in df['idx']]
输出：
idx score name0 (3,) 0.773 (D,)1 (3, 5) 0.841 (D, F)2 (1, 3, 5) 0.862 (B, D, F)3 (1, 3, 5, 10) 0.874 (B, D, F, K)4 (1, 3, 5, 8, 10) 0.883 (B, D, F, I, K)
<details><summary>英文:</summary>You need a few steps:- creating a mapping dictionary for the index -> value of the list,- converting the strings representation of tuples to tuples with [`ast.literal_eval`](https://docs.python.org/3/library/ast.html#ast.literal_eval),- mapping the values with a tuple comprehension
from ast import literal_eval
d = dict(enumerate(name_list))
df['name'] = [tuple(d.get(x, '?') for x in literal_eval(t))
for t in df['idx']]
If you are sure that the indices are valid, no need for the dictionary:
df['name'] = [tuple(name_list[x] for x in literal_eval(t))
for t in df['idx']]
For a string as output:
df['name'] = [f"({', '.join(tuple(name_list[x] for x in literal_eval(t)))})"
for t in df['idx']]
Output:
idx score name
0 (3,) 0.773 (D,)
1 (3, 5) 0.841 (D, F)
2 (1, 3, 5) 0.862 (B, D, F)
3 (1, 3, 5, 10) 0.874 (B, D, F, K)
4 (1, 3, 5, 8, 10) 0.883 (B, D, F, I, K)
</details># 答案2**得分**: 1这是一种使用 `str.findall()` 和 `explode()` 的方法：```pythondf.assign(name=(df['idx'].str.findall(r'\d+').explode().astype(int).map(dict(enumerate(name_list))).groupby(level=0).agg(tuple)))
输出结果：
idx score name0 (3,) 0.773 (D,)1 (3, 5) 0.841 (D, F)2 (1, 3, 5) 0.862 (B, D, F)3 (1, 3, 5, 10) 0.874 (B, D, F, K)4 (1, 3, 5, 8, 10) 0.883 (B, D, F, I, K)
英文:
Here is a way using str.findall() and explode()
df.assign(name = (df['idx'].str.findall(r'\d+').explode().astype(int).map(dict(enumerate(name_list))).groupby(level=0).agg(tuple)))
Output:
idx score name0 (3,) 0.773 (D,)1 (3, 5) 0.841 (D, F)2 (1, 3, 5) 0.862 (B, D, F)3 (1, 3, 5, 10) 0.874 (B, D, F, K)4 (1, 3, 5, 8, 10) 0.883 (B, D, F, I, K)
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