将 Dataframe 转换为列表字典

Question

我正在处理需要将 dataframe 转换为列表字典的情况。 示例 dataframe 如下：

我想将上面的 dataframe 转换为列表字典，如下所示：

dict = {"abc":[第 1 句，第 2 句]，"def"：[第 3 句]，"ghi"：[第 4 句，第 5 句]}

这是我的解决方案：

dict = {}
for idx, row in test_df.iterrows():
    if not row["label"] in dict:
        dict[row["label"]] = []
    else:
        continue

for key in dict:
    dict[key] = list()
    for idx, row in test_df.iterrows():
        if key == row["label"]:
            dict[key].append(row["sentence"])
        else:
            continue

print(dict)

我的解决方案有效，但它看起来像很多代码，应该有一个简单的出路。 有什么建议么？

Answer 1

data = pd.DataFrame([
    {"sentence": "sentence1", "label":"abc"},
    {"sentence": "sentence2", "label":"abc"},
    {"sentence": "sentence3", "label":"def"},
    {"sentence": "sentence4", "label":"ghi"},
    {"sentence": "sentence5", "label":"ghi"},
])
data

    sentence label
0  sentence1   abc
1  sentence2   abc
2  sentence3   def
3  sentence4   ghi
4  sentence5   ghi

data.groupby("label")["sentence"].apply(list).reindex().to_dict()

{'abc': ['sentence1', 'sentence2'],
 'def': ['sentence3'],
 'ghi': ['sentence4', 'sentence5']}

Answer 2

您可以使用groupby ，如下所示：

import pandas as pd

df = pd.DataFrame(
    {
        'sentence': ['sentence1', 'sentence2', 'sentence3', 'sentence4', 'sentence5'],
        'label': ['abc', 'abc', 'def', 'ghi', 'ghi']
    }
)

df = df.groupby('label')['sentence'].apply(list)

print({k: v for k, v in df.items()})

Output：

{'abc': ['sentence1', 'sentence2'], 'def': ['sentence3'], 'ghi': ['sentence4', 'sentence5']}

Answer 3

import pandas as pd

df = pd.DataFrame({'sentence':['10','20','30','40','50'], 'label' : ['abc', 'abc', 'def', 'ghi', 'ghi']})
d = {key: list(df.where(df.label == key).sentence.dropna().values) for key in set(df.label)}

使用 dict 理解