要列出的字符串向量 python

Question

我在 Python 工作，我在数据框中有一列是字符串，看起来像这样：

df['set'] 

0  [911,3040]
1  [130055, 99832, 62131]
2  [19397, 3987, 5330, 14781]
3  [76514, 70178, 70301, 76545]
4  [79185, 38367, 131155, 79433]

我希望它是：

['911','3040'],['130055','99832','62131'],['19397','3987','5330','14781'],['76514',70178','70301','76545'],['79185','38367','131155','79433']

为了能够运行 Word2Vec:

model = gensim.models.Word2Vec(df['set'] , size=100)

谢谢！

Answer 1

我认为你需要：

model = gensim.models.Word2Vec([[str(y) for y in x] for x in df['set']] , size=100)

L = [[str(y) for y in x] for x in df['set']]
print (L)

[['911', '3040'],
 ['130055', '99832', '62131'], 
 ['19397', '3987', '5330', '14781'],
 ['76514', '70178', '70301', '76545'], 
 ['79185', '38367', '131155', '79433']]

Answer 2

如果您有一列字符串，我建议您查看不同的解析方式。

这是我的做法，使用 ast.literal_eval。

>>> import ast
>>> [list(map(str, x)) for x in df['set'].apply(ast.literal_eval)]

或者，使用 pd.eval -

>>> [list(map(str, x)) for x in df['set'].apply(pd.eval)]  # 100 rows or less

或者，使用 yaml.load -

>>> import yaml
>>> [list(map(str, x)) for x in df['set'].apply(yaml.load)]

[
     ['911', '3040'], 
     ['130055', '99832', '62131'], 
     ['19397', '3987', '5330', '14781'], 
     ['76514', '70178', '70301', '76545'],
     ['79185', '38367', '131155', '79433']
 ]

Answer 3

要创建新列 (str_set)，并将 set 列中的项目转换为字符串：

df["str_set"] = [[str(item) for item in df.loc[row, "set"]] for row in range(len(df["set"]))]

Answer 4

通过简单的列表理解将每个元素转换为字符串并覆盖旧列：

df['set']  = [[str(i) for i in row] for row in df['set']]

根据提供的数据执行：

data_col = [911,3040], [130055, 99832, 62131], [19397, 3987, 5330, 14781], [76514, 70178, 70301, 76545],[79185, 38367, 131155, 79433]

out = [[str(i) for i in row] for row in data_col]

out

[['911', '3040'],
 ['130055', '99832', '62131'],
 ['19397', '3987', '5330', '14781'],
 ['76514', '70178', '70301', '76545'],
 ['79185', '38367', '131155', '79433']]

不确定这是否是大数据集的最快方法，因为有很多迭代。

要列出的字符串向量 python

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