用Pandas将一行分成多组

Question

我有一组句子，我想把它们分组，这样一个组中的所有行应该共享一个特定的单词。然而，一个句子可以属于许多组，因为它有很多单词。在

所以在下面的例子中，应该有这样一个组：

• “温度”组，包括所有行（0、1、2、3和4）
• 包含第2行和第4行的“冻结”组
• 包含行0、1、2和3的“the”组
• 仅包含行0的“metal”组。在
• 数据集中每隔一个单词分组

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'

但是我的代码抛出了一个错误，如图所示。（见下文） 由于我的任务有点复杂，我知道它可能不仅仅涉及调用groupby（）。有人能帮我用熊猫做单词组吗？在

编辑通过返回tuple(sentence.split())解决错误后，我尝试打印结果，但它似乎没有做任何事情。我想它可能只是把每一行放在一个组中：

^{pr2}$

Answer 1

要修复您的TypeError，您可以将lambda更改为

lambda sentence: tuple(sentence.split())

它将返回tuple，而不是list（和tuples和散列）。在

Answer 2

我目前的解决方案使用pandas的多索引特性。我确信可以通过更有效地使用numpy来改进它，但是我相信这将比其他python-only的答案更好：

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))

Answer 3

可以使用集合集合，这样每个单词都是唯一的。首先，我们需要得到所有句子中所有单词的列表。为此，我们将单词初始化为一个空集合，然后使用列表理解在每个句子中添加每个小写单词（在拆分句子之后）。在

接下来，我们使用字典理解来构建一个由单词集中的每个单词组成的字典。值是包含每个包含该单词的句子的数据帧。这些是通过对一个函数groupby(df.sentences.str.contains(word, case=False))进行分组，然后得到该条件为True的每个组。在

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',
 ...

要获取每个单词的索引值词典：

^{pr2}$

或者布尔指标矩阵。在

>>> [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],
 ...