两篇文章掌握Python语法和内置函数功能(第二篇)
Overview
这篇文章是接着上篇文章写的,主要介绍Python的字典结构,操作文件,正则表达式。上篇文章链接如下:
http://blog.csdn.net/xlinsist/article/details/50866079
字典
先看看下面的例子:
## 构建一个字典
dict = {}
## 向字典中存入键值对
dict['a'] = 'alpha'
dict['g'] = 'gamma'
dict['o'] = 'omega'
## {'a': 'alpha', 'o': 'omega', 'g': 'gamma'}
print dict
## Simple lookup, returns 'alpha'
print dict['a']
## Put new key/value into dict
dict['a'] = 6
'a' in dict ## True
## print dict['z'] ## Throws KeyError
if 'z' in dict: print dict['z'] ## Avoid KeyError
print dict.get('z') ## None (instead of KeyError)
## 如果没有这个Key值,将返回xlinsist
print dict.get('z', 'xlinsist')
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由于字符串和元组都是不可变的,它们可以做为key,数字也可以作为key,任何类型都可以作为一个值。
接下来,介绍几种遍历字典的方法:
## 这个key是以随机的顺序遍历的
for key in dict: print key
## 与上面方法相同,都是遍历key
for key in dict.keys(): print key
## 返回一个包含所有key的列表
dict.keys()
## 返回一个包含所有value的列表
dict.values()
## .items() is the dict expressed as (key, value) tuples
print dict.items() ## [('a', 'alpha'), ('o', 'omega'), ('g', 'gamma')]
for k, v in dict.items(): print k, '-->', v
## a --> xlnb
## o --> omega
## g --> gamma
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格式化字典:
hash = {}
hash['count'] = 42
hash['word'] = 'garfield'
# %d 代表 int, %s 代表 string
s = 'I want %(count)d copies of %(word)s' % hash
# 'I want 42 copies of garfield'
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del操作符:
# 移除定义的变量
var a = 5
del a
# 移除列表中的元素
list = ['a', 'b', 'c', 'd']
del list[0] ## Delete first element
del list[-2:] ## Delete last two elements
print list ## ['b']
# 移除字典中的元素
dict = {'a':1, 'b':2, 'c':3}
del dict['b'] ## Delete 'b' entry
print dict ## {'a':1, 'c':3}
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文件
open()函数能打开文件并返回一个文件指针。
# 'r'读、'w'写、'a'追加
f = open('foo.txt', 'rU')
for line in f:
print line
# 把整个文件读取到内存中并返回以文件的行作为元素的列表
f.readlines()
# 读取整个文件作为一个字符串
f.read()
# 向文件中写入字符串
f.write(string)
f.close()
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如果不确定某文本文件会用什么样的换行符,可以将open的第二个参数设定为 ‘rU’,指定通用换行符转化。这让你可以*地在Windows、UNIX(包括Mac OS X),以及其他的老Macintosh平台上交换文件,完全不用担心任何问题:无论你的代码在什么平台上运行,各种换行符都被映射成 ‘\n’。
正则表达式
基本的模式
- . (点) – 匹配任何单个字符除了换行符 ‘\n’
- \w – 匹配数字和字母。\W取反
- \s – 匹配空白字符(空格、换行、制表符…)。\S取反
- \t, \n, \r – tab, newline, return
- \d – 匹配数字
- + – 出现1次或更多次
- * – 出现0次或更多次
- ? – 出现0次或1次
来个例子:
## i+ = one or more i's, as many as possible.
match = re.search(r'pi+', 'piiig') => found, match.group() == "piii"
## Finds the first/leftmost solution, and within it drives the +
## as far as possible (aka 'leftmost and largest').
## In this example, note that it does not get to the second set of i's.
match = re.search(r'i+', 'piigiiii') => found, match.group() == "ii"
## \s* = zero or more whitespace chars
## Here look for 3 digits, possibly separated by whitespace.
match = re.search(r'\d\s*\d\s*\d', 'xx1 2 3xx') => found, match.group() == "1 2 3"
match = re.search(r'\d\s*\d\s*\d', 'xx12 3xx') => found, match.group() == "12 3"
match = re.search(r'\d\s*\d\s*\d', 'xx123xx') => found, match.group() == "123"
## ^ = matches the start of string, so this fails:
match = re.search(r'^b\w+', 'foobar') => not found, match == None
## but without the ^ it succeeds:
match = re.search(r'b\w+', 'foobar') => found, match.group() == "bar"
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模式串前的字母r指定python字符串为原始串。先看下面的例子:
由上图可以看出,加r的字符串就是2个字符:’\’和’n’;没加r的字符串python解释器认为它就是一个换行符,所以长度为1。在看下面的例子:
str = r'\n'
match = re.search(r'\n', str)
if match:
print 'found', match.group() ## 'found word:cat'
else:
print 'did not find'
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上面代码的输出结果为’did not find’,这是因为str为包含’\’和’n’的两个字符,而正则系统将’\’和’n’相邻认做是换行符,所以并不匹配。但是如果我把上面的str=r’\n’改为str = ‘\n’,那么输出结果为found,\n(这个我们并不一定看见这个字符串,我们会看到控制台上出现一行空白)。
其实我们可以认为正则和python是两个不同的编程语言,对待字符串的方式不相同。更加具体的解释参考:http://*.com/questions/12871066/what-exactly-is-a-raw-string-regex-and-how-can-you-use-it
方括号
[abc] 匹配 ‘a’ or ‘b’ or ‘c’.
## 方括号中的点就是字面上的意义,并没有什么特殊的含义
match = re.search(r'[\w.-]aaa@qq.com[\w.-]+', 'purple aaa@qq.com monkey dishwasher')
if match:
print match.group() ## 'aaa@qq.com'
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[^ab] 匹配除了 ‘a’ or ‘b’的任何字符。
用括号做组抽取
str = 'purple aaa@qq.com monkey dishwasher'
match = re.search('([\w.-]+)@([\w.-]+)', str)
if match:
print match.group() ## 'aaa@qq.com' (the whole match)
print match.group(1) ## 'alice-b' (the username, group 1)
print match.group(2) ## 'google.com' (the host, group 2)
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findall()函数找到字符串中的所有匹配项,并以字符串列表的形式返回。
## Suppose we have a text with many email addresses
str = 'purple aaa@qq.com, blah monkey aaa@qq.com blah dishwasher'
## Here re.findall() returns a list of all the found email strings
emails = re.findall(r'[\w\.-]aaa@qq.com[\w\.-]+', str) ## ['aaa@qq.com', 'aaa@qq.com']
for email in emails:
# do something with each found email string
print email
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str = 'purple aaa@qq.com, blah monkey aaa@qq.com blah dishwasher'
tuples = re.findall(r'([\w\.-]+)@([\w\.-]+)', str)
print tuples ## [('alice', 'google.com'), ('bob', 'abc.com')]
for tuple in tuples:
print tuple[0] ## username
print tuple[1] ## host
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以上内容整理自Google’s Python Class