从列表中删除标点符号

Question

我正在研究“独立宣言”的样本，并计算其中词语长度的频率。

文件中的示例文本：

"When in the Course of human events it becomes necessary for one people to dissolve the political bands which have connected them with another and to assume among the powers of the earth, the separate and equal station to which the Laws of Nature and of Nature's God entitle them, a decent respect to the opinions of mankind requires 
that they should declare the causes which impel them to the separation."

注意：字长不能包含任何标点符号，例如来自string.punctuation的任何内容。

预期结果（样本）：

Length Count
1 16
2 267
3 267
4 169
5 140
6 112
7 99
8 68
9 61
10 56
11 35
12 13
13 9
14 7
15 2

我目前仍然在从已转换为列表的文件中删除标点符号。

这是我到目前为止所尝试的内容：

import sys
import string

def format_text(fname):
        punc = set(string.punctuation)
        words = fname.read().split()
        return ''.join(word for word in words if word not in punc)

try:
    with open(sys.argv[1], 'r') as file_arg:
        file_arg.read()
except IndexError:
    print('You need to provide a filename as an arguement.')
    sys.exit()

fname = open(sys.argv[1], 'r')
formatted_text = format_text(fname)
print(formatted_text)

Answer 1

您可以从单词中删除标点符号，并避免将所有文件读入内存：

punc = string.punctuation
return ' '.join(word.strip(punc) for line in fname for word in line.split())

如果您要从'删除Nature's，则需要翻译：

from string import punctuation

# use ord of characters you want to replace as keys and what you want to replace them with as values
tbl = {ord(k):"" for k in punctuation}
return ' '.join(line.translate(tbl) for line in fname)

要获得频率，请使用Counter dict：

from collections import Counter
freq = Counter(len(word.translate(tbl)) for line in fname for word in line.split())

或者取决于你的方法：

freq = Counter(len(word.strip(punc)) for line in fname for word in line.split())

使用上述问题中的行作为示例：

lines =""""When in the Course of human events it becomes necessary for one people to dissolve the political bands which have connected them with another and to assume among the powers of the earth, the separate and equal station to which the Laws of Nature and of Nature's God entitle them, a decent respect to the opinions of mankind requires
that they should declare the causes which impel them to the separation."""

from collections import Counter
freq = Counter(len(word.strip(punctuation)) for line in lines.splitlines() for word in line.split())
print(freq.most_common()) 

输出键/值对的元组，从单词长度开始，一直到最低，键是长度，第二个元素是频率：

[(3, 15), (2, 12), (4, 9), (5, 9), (6, 9), (7, 7), (8, 5), (9, 3), (1, 1), (10, 1)]

如果您想从1个字母单词开始输出频率而不按顺序排序：

mx = max(freq.values())
for i in range(1, mx+1):
    v = freq[i]
    if v:
        print("length {} words appeared {} time/s.".format(i, v) )

输出：

length 1 words appeared 1 time/s.
length 2 words appeared 12 time/s.
length 3 words appeared 15 time/s.
length 4 words appeared 9 time/s.
length 5 words appeared 9 time/s.
length 6 words appeared 9 time/s.
length 7 words appeared 7 time/s.
length 8 words appeared 5 time/s.
length 9 words appeared 3 time/s.
length 10 words appeared 1 time/s.

对于丢失的键，Counter dict与普通dict不同，不会返回keyError，但会返回0的值，因此if v对于文件中出现的字长只会为True。

如果要打印已清理的数据，将所有逻辑放在功能中：

def clean_text(fname):
    punc = string.punctuation
    return [word.strip(punc) for line in fname for word in line.split()]


def get_freq(cleaned):
    return Counter(len(word) for word in cleaned)


def freq_output(d):
    mx = max(d.values())
    for i in range(1, mx + 1):
        v = d[i]
        if v:
            print("length {} words appeared {} time/s.".format(i, v))

try:
    with open(sys.argv[1], 'r') as file_arg:
        file_arg.read()
except IndexError:
    print('You need to provide a filename as an arguement.')
    sys.exit()

fname = open(sys.argv[1], 'r')
formatted_text = clean_text(fname)

print(" ".join(formatted_text))
print()
freq = get_freq(formatted_text)

freq_output(freq) 

在你的问题片段输出上运行：

~$ python test.py test.txt
When in the Course of human events it becomes necessary for one people  
to dissolve the political bands which have connected them with another
and to assume among the powers of the earth the separate and equal station 
 to which the Laws of Nature and of Nature's God entitle them a decent 
respect to the opinions of mankind requires that they should declare 
the causes which impel them to the separation

length 1 words appeared 1 time/s.
length 2 words appeared 12 time/s.
length 3 words appeared 15 time/s.
length 4 words appeared 9 time/s.
length 5 words appeared 9 time/s.
length 6 words appeared 9 time/s.
length 7 words appeared 7 time/s.
length 8 words appeared 5 time/s.
length 9 words appeared 3 time/s.
length 10 words appeared 1 time/s.

如果您只关心频率输出，请一次性完成所有操作：

import sys
import string


def freq_output(fname):
    from string import punctuation

    tbl = {ord(k): "" for k in punctuation}
    d = Counter(len(word.strip(punctuation)) for line in fname for word in line.split())
    d = Counter(len(word.translate(tbl)) for line in fname for word in line.split())
    mx = max(d.values())
    for i in range(1, mx + 1):
        v = d[i]
        if v:
            print("length {} words appeared {} time/s.".format(i, v))


try:
    with open(sys.argv[1], 'r') as file_arg:
        file_arg.read()
except IndexError:
    print('You need to provide a filename as an arguement.')
    sys.exit()

fname = open(sys.argv[1], 'r')

freq_output(fname)

使用适用于d的任何方法。

Answer 2

您可以使用翻译去除标点符号：

import string

words = fname.read().translate(None, string.punctuation).split()

Best way to strip punctuation from a string in Python

<强> py2.7 ：

import string
from collections import defaultdict
from collections import Counter

def s1():
    with open("myfile.txt", "r") as f:
        counts = defaultdict(int)
        for line in f:
            words = line.translate(None, string.punctuation).split()
            for length in map(len, words):
                counts[length] += 1
    return counts

def s2():
    with open("myfile.txt", "r") as f:
        counts = Counter(length for line in f for length in map(len, line.translate(None, string.punctuation).split()))
    return counts

print s1()
defaultdict(<type 'int'>, {1: 111, 2: 1169, 3: 1100, 4: 1470, 5: 1425, 6: 1318, 7: 1107, 8: 875, 9: 938, 10: 108, 11: 233, 12: 146})

print s2()
Counter({4: 1470, 5: 1425, 6: 1318, 2: 1169, 7: 1107, 3: 1100, 9: 938, 8: 875, 11: 233, 12: 146, 1: 111, 10: 108})

在python 2.7中使用Counter比手动构建字典慢，因为Counter的更新方式已经实现。

%timeit s1()
100 loops, best of 3: 4.42 ms per loop

%timeit s2()
100 loops, best of 3: 9.27 ms per loop

<强> PY3 ：

我认为在python 3.2中计数器已更新，并且比手动构建计数器字典更快或更快。

另外python3的翻译变得不那么冗长：

import string
from collections import defaultdict
from collections import Counter

strip_punct = str.maketrans('','',string.punctuation)

def s1():
    with open("myfile.txt", "r") as f:
        counts = defaultdict(int)
        for line in f:
            words = line.translate(strip_punct).split()
            for length in map(len, words):
                counts[length] += 1
    return counts

def s2():
    with open("myfile.txt", "r") as f:
        counts = Counter(length for line in f for length in map(len, line.translate(strip_punct).split()))
    return counts

print(s1())
defaultdict(<class 'int'>, {1: 111, 2: 1169, 3: 1100, 4: 1470, 5: 1425, 6: 1318, 7: 1107, 8: 875, 9: 938, 10: 108, 11: 233, 12: 146})

print(s2())
Counter({4: 1470, 5: 1425, 6: 1318, 2: 1169, 7: 1107, 3: 1100, 9: 938, 8: 875, 11: 233, 12: 146, 1: 111, 10: 108})

%timeit s1()
100 loops, best of 3: 11.4 ms per loop

%timeit s2()
100 loops, best of 3: 11.2 ms per loop

Answer 3

您可以使用正则表达式：

import re

def format_text(fname, pattern):
    words = fname.read()
    return re.sub(p, '', words)

p = re.compile(r'[!&:;",.]')
fh = open('C:/Projects/ExplorePy/test.txt')
text = format_text(fh, p)

根据需要应用split（），可以细化模式。