优化代码以图表字数

Question

我刚刚完成了一个程序，它从书籍和图表中读取文字的字数，x轴是一本书的计数，y轴是第二本书的计数。它有效，但它的速度非常慢，我希望得到一些如何优化它的技巧。我认为我最关心的是创建一本字典，用于书籍之间相似的单词，以及一本书中的单词字典而不是另一本书。这个实现为程序增加了很多运行时间，我想找到一种pythonic方法来改进它。以下是代码：

import re   # regular expressions
import io
import collections
from matplotlib import pyplot as plt

# xs=[x1,x2,...,xn]
# Number of occurences of the word in book 1

# use

# ys=[y1.y2,...,yn]
# Number of occurences of the word in book 2

# plt.plot(xs,ys)
# save as svg or pdf files

word_pattern = re.compile(r'\w+')
# with version ensures closing even if there are failures
with io.open("swannsway.txt") as f:
    text = f.read() # read as a single large string
    book1 = word_pattern.findall(text)  # pull out words
    book1 = [w.lower() for w in book1 if len(w)>=3]

with io.open("moby_dick.txt") as f:
    text = f.read() # read as a single large string
    book2 = word_pattern.findall(text)  # pull out words
    book2 = [w.lower() for w in book2 if len(w)>=3]


#Convert these into relative percentages/total book length

wordcount_book1 = {}
for word in book1:
    if word in wordcount_book1:
        wordcount_book1[word]+=1
    else:
        wordcount_book1[word]=1

'''
for word in wordcount_book1:
    wordcount_book1[word] /= len(wordcount_book1)

for word in wordcount_book2:
    wordcount_book2[word] /= len(wordcount_book2)
'''

wordcount_book2 = {}
for word in book2:
    if word in wordcount_book2:
        wordcount_book2[word]+=1
    else:
        wordcount_book2[word]=1


common_words = {}

for i in wordcount_book1:
    for j in wordcount_book2:
        if i == j:
            common_words[i] = [wordcount_book1[i], wordcount_book2[j]]
            break

book_singles= {}
for i in wordcount_book1:
    if i not in common_words:
        book_singles[i] = [wordcount_book1[i], 0]
for i in wordcount_book2:
    if i not in common_words:
        book_singles[i] = [0, wordcount_book2[i]]

wordcount_book1 = collections.Counter(book1)
wordcount_book2 = collections.Counter(book2)

# how many words of different lengths?

word_length_book1 = collections.Counter([len(word) for word in book1])
word_length_book2 = collections.Counter([len(word) for word in book2])

print(wordcount_book1)

#plt.plot(list(word_length_book1.keys()),list(word_length_book1.values()), list(word_length_book2.keys()), list(word_length_book2.values()), 'bo')
for i in range(len(common_words)):
    plt.plot(list(common_words.values())[i][0], list(common_words.values())[i][1], 'bo', alpha = 0.2)
for i in range(len(book_singles)):
    plt.plot(list(book_singles.values())[i][0], list(book_singles.values())[i][1], 'ro', alpha = 0.2)
plt.ylabel('Swannsway')
plt.xlabel('Moby Dick')
plt.show()
#key:value

Answer 1

您的大部分代码只有轻微的低效率，而我试图解决这些问题。你最大的延迟是密谋book_singles，我相信我已经修好了。细节：我换了这个：

word_pattern = re.compile(r'\w+')

为：

word_pattern = re.compile(r'[a-zA-Z]{3,}')

因为book_singles足够大而且不包含数字！通过在模式中包含最小大小，我们消除了对此循环的需求：

book1 = [w.lower() for w in book1 if len(w)>=3]

和book2匹配的那个。这里：

book1 = word_pattern.findall(text)  # pull out words
book1 = [w.lower() for w in book1 if len(w)>=3]

我移动.lower()所以我们只做一次，而不是每一个字：

book1 = word_pattern.findall(text.lower())  # pull out words
book1 = [w for w in book1 if len(w) >= 3]

由于它可能在 C 中实施，这可能是一场胜利。这样：

wordcount_book1 = {}
for word in book1:
    if word in wordcount_book1:
        wordcount_book1[word]+=1
    else:
        wordcount_book1[word]=1

我已切换为使用defaultdict，因为您已经导入了集合：

wordcount_book1 = collections.defaultdict(int)
for word in book1:
    wordcount_book1[word] += 1

对于这些循环：

common_words = {}

for i in wordcount_book1:
    for j in wordcount_book2:
        if i == j:
            common_words[i] = [wordcount_book1[i], wordcount_book2[j]]
            break

book_singles= {}
for i in wordcount_book1:
    if i not in common_words:
        book_singles[i] = [wordcount_book1[i], 0]
for i in wordcount_book2:
    if i not in common_words:
        book_singles[i] = [0, wordcount_book2[i]]

我重写了第一个循环，这是一场灾难，然后让它做了双重任务，因为它已经完成了第二个循环的工作：

common_words = {}
book_singles = {}

for i in wordcount_book1:
    if i in wordcount_book2:
        common_words[i] = [wordcount_book1[i], wordcount_book2[i]]
    else:
        book_singles[i] = [wordcount_book1[i], 0]

for i in wordcount_book2:
    if i not in common_words:
        book_singles[i] = [0, wordcount_book2[i]]

最后，这些绘图循环在他们一遍又一遍地走common_words.values()和book_singles.values()的方式以及他们一次绘制一个点的方式上效率非常低：

for i in range(len(common_words)):
    plt.plot(list(common_words.values())[i][0], list(common_words.values())[i][1], 'bo', alpha = 0.2)
for i in range(len(book_singles)):
    plt.plot(list(book_singles.values())[i][0], list(book_singles.values())[i][1], 'ro', alpha = 0.2)

我将它们改为：

counts1, counts2 = zip(*common_words.values())
plt.plot(counts1, counts2, 'bo', alpha=0.2)

counts1, counts2 = zip(*book_singles.values())
plt.plot(counts1, counts2, 'ro', alpha=0.2)

完整的返工代码，它会遗漏您计算但未在示例中使用的内容：

import re  # regular expressions
import collections
from matplotlib import pyplot as plt

# xs=[x1,x2,...,xn]
# Number of occurrences of the word in book 1

# use

# ys=[y1.y2,...,yn]
# Number of occurrences of the word in book 2

# plt.plot(xs,ys)
# save as svg or pdf files

word_pattern = re.compile(r'[a-zA-Z]{3,}')

# with ensures closing of file even if there are failures
with open("swannsway.txt") as f:
    text = f.read() # read as a single large string
    book1 = word_pattern.findall(text.lower())  # pull out words

with open("moby_dick.txt") as f:
    text = f.read() # read as a single large string
    book2 = word_pattern.findall(text.lower())  # pull out words

# Convert these into relative percentages/total book length

wordcount_book1 = collections.defaultdict(int)
for word in book1:
    wordcount_book1[word] += 1

wordcount_book2 = collections.defaultdict(int)
for word in book2:
    wordcount_book2[word] += 1

common_words = {}
book_singles = {}

for i in wordcount_book1:
    if i in wordcount_book2:
        common_words[i] = [wordcount_book1[i], wordcount_book2[i]]
    else:
        book_singles[i] = [wordcount_book1[i], 0]

for i in wordcount_book2:
    if i not in common_words:
        book_singles[i] = [0, wordcount_book2[i]]

counts1, counts2 = zip(*common_words.values())
plt.plot(counts1, counts2, 'bo', alpha=0.2)

counts1, counts2 = zip(*book_singles.values())
plt.plot(counts1, counts2, 'ro', alpha=0.2)

plt.xlabel('Moby Dick')
plt.ylabel('Swannsway')
plt.show()

<强>输出

您可以删除stop words以减少高分词并带出有趣的数据。

Answer 2

以下是优化代码的一些提示。

计算单词的出现次数。 使用Counter库中的collections课程（请参阅this post）：

from collections import Counter
wordcount_book1 = Counter(book1)
wordcount_book2 = Counter(book2)

查找常用且独特的字词。 使用set课程。所有单词都是联合，常用单词是交集，唯一单词就是区别。

book1_words = set(wordcount_book1.keys())
book2_words = set(wordcount_book2.keys())
all_words = book1_words | book2_words 
common_words = book1_words & book2_words
book_singles = [book1_words - common_words, book2_words - common_words]

计算单词长度。 首先计算所有单词的长度，然后将其乘以每本书的单词数：

word_length = Counter([len(w) for w in all_words])
word_length_book1 = {w:word_length[w]*wordcount_book1[w] for w in book1_words})
word_length_book1 = {w:word_length[w]*wordcount_book2[w] for w in book2_words}

可能这些情节应该没有循环，但不幸的是我不明白你想要绘制的是什么。