从数据帧创建句子(行)至POS标签计数(列)矩阵 [英] Create sentence (row) to POS tags counts (column) matrix from a dataframe
本文介绍了从数据帧创建句子(行)至POS标签计数(列)矩阵的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我正在尝试建立一个矩阵，其中第一行将成为语音的一部分，第一列将成为句子.矩阵中的值应显示句子中此类POS的数量.
I am trying to build a matrix where the first row will be a part of speech, first column a sentence. values in the matrix should show the number of such POS in a sentence.
所以我以这种方式创建POS标签:
So I am creating POS tags in this way:
data = pd.read_csv(open('myfile.csv'),sep=';')
target = data["label"]
del data["label"]
data.sentence = data.sentence.str.lower() # All strings in data frame to lowercase
for line in data.sentence:
Line_new= nltk.pos_tag(nltk.word_tokenize(line))
print(Line_new)
输出为:
[('together', 'RB'), ('with', 'IN'), ('the', 'DT'), ('6th', 'CD'), ('battalion', 'NN'), ('of', 'IN'), ('the', 'DT')]
如何从这样的输出中创建一个上面已经描述过的矩阵?
How can I create a matrix which I have described above from such output?
更新: 所需的输出是
NN VB IN VBZ DT
I was there 1 1 1 0 0
He came there 0 0 1 1 1
myfile.csv:
myfile.csv:
"A child who is exclusively or predominantly oral (using speech for communication) can experience social isolation from his or her hearing peers, particularly if no one takes the time to explicitly teach them social skills that other children acquire independently by virtue of having normal hearing.";"certain"
"Preliminary Discourse to the Encyclopedia of Diderot";"certain"
"d'Alembert claims that it would be ignorant to perceive that everything could be known about a particular subject.";"certain"
"However, as the overemphasis on parental influence of psychodynamics theory has been strongly criticized in the previous century, modern psychologists adopted interracial contact as a more important determinant than childhood experience on shaping people’s prejudice traits (Stephan & Rosenfield, 1978).";"uncertain"
"this can also be summarized as a distinguish behaviour on the peronnel level";"uncertain"
推荐答案
详细:
首先让我们向您的csv添加一些标头，以便在访问列时更加人性化:
First let's add some headers to your csv so that it's more humanly-readable when accessing the columns:
>>> import pandas as pd
>>> df = pd.read_csv('myfile.csv', delimiter=';')
>>> df.columns = ['sent', 'tag']
>>> df['sent']
0 Preliminary Discourse to the Encyclopedia of D...
1 d'Alembert claims that it would be ignorant to...
2 However, as the overemphasis on parental influ...
3 this can also be summarized as a distinguish b...
Name: sent, dtype: object
>>> df['tag']
0 certain
1 certain
2 uncertain
3 uncertain
现在，我们创建一个以链接方式执行word_tokenize和pos_tag的函数tok_and_tag:
Now let's create a function tok_and_tag that does word_tokenize and pos_tag in a chained manner:
>>> from nltk import word_tokenize, pos_tag
>>> from functools import partial
>>> tok_and_tag = lambda x: pos_tag(word_tokenize(x))
>>> df['sent'][0]
'Preliminary Discourse to the Encyclopedia of Diderot'
>>> tok_and_tag(df['sent'][0])
[('Preliminary', 'JJ'), ('Discourse', 'NNP'), ('to', 'TO'), ('the', 'DT'), ('Encyclopedia', 'NNP'), ('of', 'IN'), ('Diderot', 'NNP')]
然后，我们可以使用 df.apply 标记并标记数据框的句子列:
Then, we can use df.apply to tokenize and tag the sentence column of the dataframe:
>>> df['sent'].apply(tok_and_tag)
0 [(Preliminary, JJ), (Discourse, NNP), (to, TO)...
1 [(d'Alembert, NN), (claims, NNS), (that, IN), ...
2 [(However, RB), (,, ,), (as, IN), (the, DT), (...
3 [(this, DT), (can, MD), (also, RB), (be, VB), ...
Name: sent, dtype: object
如果您希望句子为小写:
If you want the sentences to be lowercased:
>>> df['sent'].apply(str.lower)
0 preliminary discourse to the encyclopedia of d...
1 d'alembert claims that it would be ignorant to...
2 however, as the overemphasis on parental influ...
3 this can also be summarized as a distinguish b...
Name: sent, dtype: object
>>> df['lower_sent'] = df['sent'].apply(str.lower)
>>> df['lower_sent'].apply(tok_and_tag)
0 [(preliminary, JJ), (discourse, NN), (to, TO),...
1 [(d'alembert, NN), (claims, NNS), (that, IN), ...
2 [(however, RB), (,, ,), (as, IN), (the, DT), (...
3 [(this, DT), (can, MD), (also, RB), (be, VB), ...
Name: lower_sent, dtype: object
此外，我们需要某种方式来获取POS词汇，我们可以使用collections.Counter和itertools.chain来使列表列表变平:
Additionally, we need some sort of way to get the POS vocabulary, we can do use collections.Counter and itertools.chain to flatten the list of list:
>>> df['lower_sent']
0 preliminary discourse to the encyclopedia of d...
1 d'alembert claims that it would be ignorant to...
2 however, as the overemphasis on parental influ...
3 this can also be summarized as a distinguish b...
Name: lower_sent, dtype: object
>>> df['lower_sent'].apply(tok_and_tag)
0 [(preliminary, JJ), (discourse, NN), (to, TO),...
1 [(d'alembert, NN), (claims, NNS), (that, IN), ...
2 [(however, RB), (,, ,), (as, IN), (the, DT), (...
3 [(this, DT), (can, MD), (also, RB), (be, VB), ...
Name: lower_sent, dtype: object
>>> df['tagged_sent'] = df['lower_sent'].apply(tok_and_tag)
>>> tokens, tags = zip(*chain(*df['tagged_sent'].tolist()))
>>> tags
('JJ', 'NN', 'TO', 'DT', 'NN', 'IN', 'NN', 'NN', 'NNS', 'IN', 'PRP', 'MD', 'VB', 'JJ', 'TO', 'VB', 'IN', 'NN', 'MD', 'VB', 'VBN', 'IN', 'DT', 'JJ', 'NN', '.', 'RB', ',', 'IN', 'DT', 'NN', 'IN', 'JJ', 'NN', 'IN', 'NNS', 'NN', 'VBZ', 'VBN', 'RB', 'VBN', 'IN', 'DT', 'JJ', 'NN', ',', 'JJ', 'NNS', 'VBD', 'JJ', 'NN', 'IN', 'DT', 'RBR', 'JJ', 'NN', 'IN', 'NN', 'NN', 'IN', 'VBG', 'JJ', 'NN', 'NNS', '(', 'NN', 'CC', 'NN', ',', 'CD', ')', '.', 'DT', 'MD', 'RB', 'VB', 'VBN', 'IN', 'DT', 'JJ', 'NN', 'IN', 'DT', 'NNS', 'NN')
>>> set(tags)
{'CC', 'VB', ')', 'NNS', ',', 'JJ', 'VBZ', 'DT', 'NN', 'PRP', 'RBR', 'TO', 'VBD', '(', 'VBN', '.', 'MD', 'IN', 'RB', 'VBG', 'CD'}
>>> possible_tags = sorted(set(tags))
>>> possible_tags
['(', ')', ',', '.', 'CC', 'CD', 'DT', 'IN', 'JJ', 'MD', 'NN', 'NNS', 'PRP', 'RB', 'RBR', 'TO', 'VB', 'VBD', 'VBG', 'VBN', 'VBZ']
>>> possible_tags_counter = Counter({p:0 for p in possible_tags})
>>> possible_tags_counter
Counter({'NNS': 0, 'VBZ': 0, 'DT': 0, '(': 0, 'JJ': 0, 'VBD': 0, ')': 0, 'RB': 0, 'VBG': 0, 'RBR': 0, 'VB': 0, 'IN': 0, 'CC': 0, ',': 0, 'PRP': 0, 'CD': 0, 'VBN': 0, '.': 0, 'MD': 0, 'NN': 0, 'TO': 0})
要遍历每个带标记的句子并获得POS的计数:
To iterate through each tagged sentence and get the counts of POS:
>>> df['tagged_sent'].apply(lambda x: Counter(list(zip(*x))[1]))
0 {'NN': 3, 'IN': 1, 'TO': 1, 'DT': 1, 'JJ': 1}
1 {'NN': 3, 'VB': 3, 'PRP': 1, 'TO': 1, 'DT': 1,...
2 {')': 1, 'JJ': 6, 'NN': 11, 'CC': 1, 'NNS': 3,...
3 {'DT': 3, 'VB': 1, 'NN': 2, 'VBN': 1, 'NNS': 1...
Name: tagged_sent, dtype: object
>>> df['pos_counts'] = df['tagged_sent'].apply(lambda x: Counter(list(zip(*x))[1]))
>>> df['pos_counts']
0 {'NN': 3, 'IN': 1, 'TO': 1, 'DT': 1, 'JJ': 1}
1 {'NN': 3, 'VB': 3, 'PRP': 1, 'TO': 1, 'DT': 1,...
2 {')': 1, 'JJ': 6, 'NN': 11, 'CC': 1, 'NNS': 3,...
3 {'DT': 3, 'VB': 1, 'NN': 2, 'VBN': 1, 'NNS': 1...
Name: pos_counts, dtype: object
# Now we can add in the POS that don't appears in the sentence with 0 counts:
>>> def add_pos_with_zero_counts(counter, keys_to_add):
... for k in keys_to_add:
... counter[k] = counter.get(k, 0)
... return counter
...
>>> df['pos_counts'].apply(lambda x: add_pos_with_zero_counts(x, possible_tags))
0 {'VB': 0, 'IN': 1, 'PRP': 0, 'DT': 1, 'CC': 0,...
1 {'VB': 3, ')': 0, 'DT': 1, 'CC': 0, 'RB': 0, '...
2 {'VB': 0, ')': 1, 'JJ': 6, 'NN': 11, 'CC': 1, ...
3 {'VB': 1, 'IN': 2, 'PRP': 0, 'NN': 2, 'CC': 0,...
Name: pos_counts, dtype: object
>>> df['pos_counts_with_zero'] = df['pos_counts'].apply(lambda x: add_pos_with_zero_counts(x, possible_tags))
现在将这些值展平到列表中:
Now flatten the values into the list:
>>> df['pos_counts_with_zero'].apply(lambda x: [count for tag, count in sorted(x.most_common())])
0 [0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 3, 0, 0, 0, 0, ...
1 [0, 0, 0, 1, 0, 0, 1, 3, 2, 2, 3, 1, 1, 0, 0, ...
2 [1, 1, 3, 1, 1, 1, 3, 7, 6, 0, 11, 3, 0, 2, 1,...
3 [0, 0, 0, 0, 0, 0, 3, 2, 1, 1, 2, 1, 0, 1, 0, ...
Name: pos_counts_with_zero, dtype: object
>>> df['sent_vector'] = df['pos_counts_with_zero'].apply(lambda x: [count for tag, count in sorted(x.most_common())])
现在，您需要创建一个新的矩阵来存储BoW:
Now you need to create a new matrix to store the BoW:
>>> df2 = pd.DataFrame(df['sent_vector'].tolist)
>>> df2.columns = sorted(possible_tags)
瞧，
>>> df2
( ) , . CC CD DT IN JJ MD ... NNS PRP RB RBR TO VB VBD \
0 0 0 0 0 0 0 1 1 1 0 ... 0 0 0 0 1 0 0
1 0 0 0 1 0 0 1 3 2 2 ... 1 1 0 0 1 3 0
2 1 1 3 1 1 1 3 7 6 0 ... 3 0 2 1 0 0 1
3 0 0 0 0 0 0 3 2 1 1 ... 1 0 1 0 0 1 0
VBG VBN VBZ
0 0 0 0
1 0 1 0
2 1 2 1
3 0 1 0
[4 rows x 21 columns]
简而言之:
from collections import Counter
from itertools import chain
import pandas as pd
from nltk import word_tokenize, pos_tag
df = pd.read_csv('myfile.csv', delimiter=';')
df.columns = ['sent', 'tag']
tok_and_tag = lambda x: pos_tag(word_tokenize(x))
df['lower_sent'] = df['sent'].apply(str.lower)
df['tagged_sent'] = df['lower_sent'].apply(tok_and_tag)
possible_tags = sorted(set(list(zip(*chain(*df['tagged_sent'])))[1]))
def add_pos_with_zero_counts(counter, keys_to_add):
for k in keys_to_add:
counter[k] = counter.get(k, 0)
return counter
# Detailed steps.
df['pos_counts'] = df['tagged_sent'].apply(lambda x: Counter(list(zip(*x))[1]))
df['pos_counts_with_zero'] = df['pos_counts'].apply(lambda x: add_pos_with_zero_counts(x, possible_tags))
df['sent_vector'] = df['pos_counts_with_zero'].apply(lambda x: [count for tag, count in sorted(x.most_common())])
# All in one.
df['sent_vector'] = df['tagged_sent'].apply(lambda x:
[count for tag, count in sorted(
add_pos_with_zero_counts(
Counter(list(zip(*x))[1]),
possible_tags).most_common()
)
]
)
df2 = pd.DataFrame(df['sent_vector'].tolist())
df2.columns = possible_tags
这篇关于从数据帧创建句子(行)至POS标签计数(列)矩阵的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
查看全文
