Question

我有一组句子，我想将它们分组，这样组中的所有行都应该共享一个特定的单词。但是，一个句子可以属于许多组，因为它中有很多单词。

所以在下面的例子中，应该有这样的组：

A＆＃39;温度＆＃39;包含所有行（0,1,2,3和4）的组
A＆＃39;冻结＆＃39;包括第2行和第4行的组
A＆＃39;＆＃39;包含行0,1,2和3的组
A＆＃39;金属＆＃39;仅包含第0行的组。
数据集中每个其他单词的分组

import pandas as pd

# An example data set
df = pd.DataFrame({"sentences": [
    "two long pieces of metal fixed together, each of which bends a different amount when they are both heated to the same temperature",
    "the temperature at which a liquid boils",
    "a system for measuring temperature that is part of the metric system, in which water freezes at 0 degrees and boils at 100 degrees",
    "a unit for measuring temperature. Measurements are often expressed as a number followed by the symbol °",
    "a system for measuring temperature in which water freezes at 32º and boils at 212º"
]})

# Create a new series which is a list of words in each "sentences" column
df['words'] = df['sentences'].apply(lambda sentence: sentence.split(" "))

# Try to group by this new column 
df.groupby('words').count()

# TypeError: unhashable type: 'list'

~~但是我的代码会抛出错误，如图所示。~~（见下文）由于我的任务有点复杂，我知道它可能涉及的不仅仅是调用groupby（）。有人可以帮我用熊猫制作单词组吗？

编辑通过返回tuple(sentence.split())（感谢ethan-furman）解决错误后，我尝试打印结果，但它似乎没有做任何事情。我认为它可能只是将每一行放在一个组中：

print(df.groupby('words').count())

# sentences    5
# dtype: int64

Answer 1

您可以使用集合集合，以便每个单词都是唯一的。首先，我们需要获得所有句子中所有单词的列表。为此，我们将单词初始化为空集，然后使用列表解析在每个句子中添加每个小写单词（在分割句子之后）。

接下来，我们使用字典理解来构建一个字典集中每个单词的字典。该值是包含包含该单词的每个句子的数据框。这些是通过对函数groupby(df.sentences.str.contains(word, case=False))进行分组，然后使每个组的条件为True得到的。

words = set()
_ = [words.add(word.lower()) for sentence in df.sentences for word in sentence.split()]

word_dict = {word: df.groupby(df.sentences.str.contains(word, case=False)).get_group(True) 
             for word in words}

>>> word_dict['temperature']
                                           sentences
0  two long pieces of metal fixed together, each ...
1            the temperature at which a liquid boils
2  a system for measuring temperature that is par...
3  a unit for measuring temperature. Measurements...
4  a system for measuring temperature in which wa...

>>> word_dict['freezes']
                                           sentences
2  a system for measuring temperature that is par...
4  a system for measuring temperature in which wa...

>>> words
{'0',
 '100',
 '212\xc2\xba',
 '32\xc2\xba',
 'a',
 'amount',
 'and',
 'are',
 'as',
 'at',
 'bends',
 ...

获取每个单词的索引值字典：

>>> {word: word_dict[word].index.tolist() for word in word_dict}
{'0': [2],
 '100': [2],
 '212\xc2\xba': [4],
 '32\xc2\xba': [4],
 'a': [0, 1, 2, 3, 4],
 'amount': [0],
 'and': [2, 4],
 'are': [0, 3],
 'as': [2, 3, 4],
 'at': [0, 1, 2, 3, 4],
 'bends': [0],
 'boils': [1, 2, 4],
 'both': [0],
 'by': [3],
 'degrees': [2],
 'different': [0],
 'each': [0],
 'expressed': [3],
 'fixed': [0],
 'followed': [3],
 'for': [2, 3, 4],
 'freezes': [2, 4],
 ...

或布尔指示符矩阵。

>>> [df.sentences.str.contains(word, case='lower').tolist() for word in word_dict]
[[False, False, True, False, True],
 [False, False, False, True, False],
 [True, False, False, False, False],
 [False, False, True, False, False],
 ...

Answer 2

要修复TypeError，您可以将lambda更改为

lambda sentence: tuple(sentence.split())

将返回tuple而不是list（以及tuples并且可以播放）。

Answer 3

我目前的解决方案使用pandas的MultiIndex功能。我相信它可以通过更有效地使用numpy来改进，但我相信这会比其他只有python的答案明显更好：

import pandas as pd
import numpy as np

# An example data set
df = pd.DataFrame({"sentences": [
    "two long pieces of metal fixed together, each of which bends a different amount when they are both heated to the same temperature",
    "the temperature at which a liquid boils",
    "a system for measuring temperature that is part of the metric system, in which water freezes at 0 degrees and boils at 100 degrees",
    "a unit for measuring temperature. Measurements are often expressed as a number followed by the symbol °",
    "a system for measuring temperature in which water freezes at 32º and boils at 212º"
]})

# Create a new series which is a list of words in each "sentences" column
df['words'] = df['sentences'].apply(lambda sentence: sentence.split(" "))

# This is all the words in the dataset. Each word will be its own index (level of the MultiIndex)
names = np.unique(df['words'].sum())

# Create an array of tuples, one tuple for each row of data
# Each tuple contains True if the row has that word in it, and False if it does not
values = df['words'].map(
    lambda words: np.vectorize(
        lambda word:
            True if word in words else False)(names)
)

# Make a multindex
index = pd.MultiIndex.from_tuples(values, names=names)

# Add the MultiIndex without creating a new data frame
df.set_index(index, inplace=True)

# Find all the rows that have the word 'temperature'
xs = df.xs(True, level='temperature')

print(xs.to_string(index=False))

使用pandas将行分组到多个组中

3 个答案: