如何顯示包含這些字符的所有單詞？

Question

我有一個文本文件，我想顯示包含z和x字符的所有單詞。

我該怎麼做？

Answer 1

如果你不想有2個問題：

for word in file('myfile.txt').read().split(): 
    if 'x' in word and 'z' in word: 
     print word 

Answer 2

聽起來像是Regular Expressions的工作。閱讀並嘗試一下。如果遇到問題，請更新您的問題，我們可以幫助您解決具體問題。

Answer 3

假設你擁有整個文件在內存中一個大字符串，這一個詞的定義是「字母的連續序列」，那麼你可以做這樣的事情：

import re 
for word in re.findall(r"\w+", mystring): 
    if 'x' in word and 'z' in word: 
     print word 

Answer 4

>>> import re 
>>> pattern = re.compile('\b(\w*z\w*x\w*|\w*x\w*z\w*)\b') 
>>> document = '''Here is some data that needs 
... to be searched for words that contain both z 
... and x. Blah xz zx blah jal akle asdke asdxskz 
... zlkxlk blah bleh foo bar''' 
>>> print pattern.findall(document) 
['xz', 'zx', 'asdxskz', 'zlkxlk'] 

Answer 5

>>> import re 
>>> print re.findall('(\w*x\w*z\w*|\w*z\w*x\w*)', 'axbzc azb axb abc axzb') 
['axbzc', 'axzb'] 

Answer 6

我不知道該發電機的性能，但對我來說ŧ他是這樣的：

from __future__ import print_function 
import string 

bookfile = '11.txt' # Alice in Wonderland 
hunted = 'az' # in your case xz but there is none of those in this book 

with open(bookfile) as thebook: 
    # read text of book and split from white space 
    print('\n'.join(set(word.lower().strip(string.punctuation) 
        for word in thebook.read().split() 
        if all(c in word.lower() for c in hunted)))) 
""" Output: 
zealand 
crazy 
grazed 
lizard's 
organized 
lazy 
zigzag 
lizard 
lazily 
gazing 
"" 

「

Answer 7

我只是想指出如何笨拙一些正則表達式可以在比較簡單的string methods-based solution provided by Wooble。

讓我們來做一些時間安排吧？

#!/usr/bin/env python 
# -*- coding: UTF-8 -*- 

import timeit 
import re 
import sys 

WORD_RE_COMPILED = re.compile(r'\w+') 
Z_RE_COMPILED = re.compile(r'(\b\w*z\w*\b)') 
XZ_RE_COMPILED = re.compile(r'\b(\w*z\w*x\w*|\w*x\w*z\w*)\b') 

########################## 
# Tim Pietzcker's solution 
# https://stackoverflow.com/questions/3962846/how-to-display-all-words-that-contain-these-characters/3962876#3962876 
# 
def xz_re_word_find(text): 
    for word in re.findall(r'\w+', text): 
     if 'x' in word and 'z' in word: 
      print word 


# Tim's solution, compiled 
def xz_re_word_compiled_find(text): 
    pattern = re.compile(r'\w+') 
    for word in pattern.findall(text): 
     if 'x' in word and 'z' in word: 
      print word 


# Tim's solution, with the RE pre-compiled so compilation doesn't get 
# included in the search time 
def xz_re_word_precompiled_find(text): 
    for word in WORD_RE_COMPILED.findall(text): 
     if 'x' in word and 'z' in word: 
      print word 


################################ 
# Steven Rumbalski's solution #1 
# (provided in the comment) 
# https://stackoverflow.com/questions/3962846/how-to-display-all-words-that-contain-these-characters/3963285#3963285 
def xz_re_z_find(text): 
    for word in re.findall(r'(\b\w*z\w*\b)', text): 
     if 'x' in word: 
      print word 


# Steven's solution #1 compiled 
def xz_re_z_compiled_find(text): 
    pattern = re.compile(r'(\b\w*z\w*\b)') 
    for word in pattern.findall(text): 
     if 'x' in word: 
      print word 


# Steven's solution #1 with the RE pre-compiled 
def xz_re_z_precompiled_find(text): 
    for word in Z_RE_COMPILED.findall(text): 
     if 'x' in word: 
      print word 


################################ 
# Steven Rumbalski's solution #2 
# https://stackoverflow.com/questions/3962846/how-to-display-all-words-that-contain-these-characters/3962934#3962934 
def xz_re_xz_find(text): 
    for word in re.findall(r'\b(\w*z\w*x\w*|\w*x\w*z\w*)\b', text): 
     print word 


# Steven's solution #2 compiled 
def xz_re_xz_compiled_find(text): 
    pattern = re.compile(r'\b(\w*z\w*x\w*|\w*x\w*z\w*)\b') 
    for word in pattern.findall(text): 
     print word 


# Steven's solution #2 pre-compiled 
def xz_re_xz_precompiled_find(text): 
    for word in XZ_RE_COMPILED.findall(text): 
     print word 


################################# 
# Wooble's simple string solution 
def xz_str_find(text): 
    for word in text.split(): 
     if 'x' in word and 'z' in word: 
      print word 


functions = [ 
     'xz_re_word_find', 
     'xz_re_word_compiled_find', 
     'xz_re_word_precompiled_find', 
     'xz_re_z_find', 
     'xz_re_z_compiled_find', 
     'xz_re_z_precompiled_find', 
     'xz_re_xz_find', 
     'xz_re_xz_compiled_find', 
     'xz_re_xz_precompiled_find', 
     'xz_str_find' 
] 

import_stuff = functions + [ 
     'text', 
     'WORD_RE_COMPILED', 
     'Z_RE_COMPILED', 
     'XZ_RE_COMPILED' 
] 


if __name__ == '__main__': 

    text = open(sys.argv[1]).read() 
    timings = {} 
    setup = 'from __main__ import ' + ','.join(import_stuff) 
    for func in functions: 
     statement = func + '(text)' 
     timer = timeit.Timer(statement, setup) 
     min_time = min(timer.repeat(3, 10)) 
     timings[func] = min_time 


    for func in functions: 
     print func + ":", timings[func], "seconds" 

運行在plaintext copy of Moby Dick這個腳本Project Gutenberg獲得的，在Python 2.6中，我得到以下計時：

xz_re_word_find: 1.21829485893 seconds 
xz_re_word_compiled_find: 1.42398715019 seconds 
xz_re_word_precompiled_find: 1.40110301971 seconds 
xz_re_z_find: 0.680151939392 seconds 
xz_re_z_compiled_find: 0.673038005829 seconds 
xz_re_z_precompiled_find: 0.673489093781 seconds 
xz_re_xz_find: 1.11700701714 seconds 
xz_re_xz_compiled_find: 1.12773990631 seconds 
xz_re_xz_precompiled_find: 1.13285303116 seconds 
xz_str_find: 0.590088844299 seconds 

在Python 3.1（使用2to3修復打印報表後），我得到以下時序：

xz_re_word_find: 2.36110496521 seconds 
xz_re_word_compiled_find: 2.34727501869 seconds 
xz_re_word_precompiled_find: 2.32607793808 seconds 
xz_re_z_find: 1.32204890251 seconds 
xz_re_z_compiled_find: 1.34104800224 seconds 
xz_re_z_precompiled_find: 1.34424304962 seconds 
xz_re_xz_find: 2.33851099014 seconds 
xz_re_xz_compiled_find: 2.29653286934 seconds 
xz_re_xz_precompiled_find: 2.32416701317 seconds 
xz_str_find: 0.656699895859 seconds 

我們可以看到，基於正則表達式的功能，往往需要兩倍的時間來的〜應變運行g是基於方法的函數，在Python 3中是超過3倍。對於一次性解析（沒有人會錯過這些毫秒），時間差異是微不足道的，但對於必須多次調用該函數的情況，基於字符串方法的方法既簡單又快捷。