从列表中删除标点符号

Question

我正在对《独立宣言》进行抽样并计算其中单词长度的频率。

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string.punctuation 中的任何内容。

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

我目前正坚持从已转换为列表的文件中删除标点符号。

这是我迄今为止尝试过的：

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

你可以用 translate 去掉标点符号：

import string

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

Best way to strip punctuation from a string in Python

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 比手动构建字典要慢，因为 Counter 的更新是实现的。

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

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

py3：

我认为在 python 3.2 中，Counter 已更新并且变得与手动构建计数器字典相同或更快。

python3 的翻译也变得不那么冗长了：

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 2

你可以去掉单词中的标点符号，也可以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())

如果您想从Nature's 中删除'，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)

要获取频率，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())

以上述问题中的行为例：

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，从最多看到的单词长度一直到最少，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)]

如果你想输出从1个字母单词开始的频率，不排序和按顺序：

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.

对于缺少的键，Counter dict 与普通 dict 不同，不会返回 keyError，但会返回值 0，因此 if v 仅对于文件中出现的字长为 True。

如果你想打印清理过的数据，把所有的逻辑放在函数中：

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 sn-p 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.

如果你只关心频率输出，那就一次性搞定：

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)

对d 使用正确的方法。

Answer 3

你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)

你喜欢用split()，模式就可以细化了。