python3 标准库一些总结(1)

 

 

1，统计个数（字符串，列表等）或者初始化字典，输出一个包含键和计数的字典或提供一个元素序列，还可以使用关键字参数讲字符串名映射到计数。

模块：collections 

构造函数： Counter 

import collections
text1 = "asbgewgrg2121aaassbsbgeeeegwwrr"
c = collections.Counter(text1)
print(c)
print(collections.Counter({'a':3,'b':2}))
print(collections.Counter(a=6, b=5))
##########输出如下#######
Counter({'g': 5, 'e': 5, 'a': 4, 's': 4, 'b': 3, 'w': 3, 'r': 3, '2': 2, '1': 2})
Counter({'a': 3, 'b': 2})
Counter({'a': 6, 'b': 5})

2，多行合并

sample_text = """
Ridiculously fast.
Django was designed to help developers take applications from concept to completion as quickly as possible.

Reassuringly secure.
Django takes security seriously and helps developers avoid many common security mistakes.

Exceedingly scalable.
Some of the busiest sites on the Web leverage Django’s ability to quickly and flexibly scale.
"""

import textwrap
print(textwrap.fill(sample_text,width=50))
#
dedented_text = textwrap.dedent(sample_text)
print('Dedented:')
print(dedented_text)
#
dedented2_text = textwrap.dedent(sample_text).strip()
for width in [45, 60]:
    print('{} Columns:\n'.format(width))
    print(textwrap.fill(dedented2_text, width=width))
    print()
#
dedented3_text = textwrap.dedent(sample_text)
wrapped = textwrap.fill(dedented3_text,width=50)
wrapped += '\n\nSecond garagraph after a blank line.'
final = textwrap.indent(wrapped,'>')
print('Quoted block:\n')
print(final)

3，re

import re

pattern = "this"
text = "Does this text match the pattern?"

match = re.search(pattern, text)
print(type(match))
s = match.start()
e = match.end()

print('Found "{}"\nin "{}"\nfrom {} to {} ("{}")'.format(match.re.pattern, match.string, s, e, text[s:e]))
#
print("#" * 20+str(2) + "#" * 20)
#2
import re
regexes = [
    re.compile(p) for p in ['this', 'that']
]
text = 'Does this text match the pattern?'
print('text: {!r}\n'.format(text))

for regex in regexes:
    print('Seeking "{}" ->'.format(regex.pattern), end=' ')
    if regex.search(text):
        print('match!')
    else:
        print("no match")

print("#" * 20+str(3) + "#" * 20)
#3
import re
text = 'abbaaabbbbaaaaa'
pattern = 'ab'
for match in re.findall(pattern,text):
    print('Found {!r}'.format(match))
print("#" * 20+str(3.2) + "#" * 20)
for match in re.finditer(pattern,text):
    s = match.start()
    e = match.end()
    print('Found {!r} at {:d}:{:d}'.format(text[s:e], s, e))
print("#" * 20 + str(4) + "#" * 20)
################输出#####################
<class 're.Match'>
Found "this"
in "Does this text match the pattern?"
from 5 to 9 ("this")
####################2####################
text: 'Does this text match the pattern?'

Seeking "this" -> match!
Seeking "that" -> no match
####################3####################
Found 'ab'
Found 'ab'
####################3.2####################
Found 'ab' at 0:2
Found 'ab' at 5:7

 4，查找单个字符串在某个序列中的计数，但不支持单个单词在string中查找，可以在列表或元祖中查找计数；

模块：collections 

构造函数： Counter 

import collections
#1
d = collections.Counter("abcdaafbbdd")
for letter in 'abcde':
    print('{} : {}'.format(letter, d[letter]))

#2
m = collections.Counter(["this","a","b"]) #()元组 ,序列
for letter in {'this'}: # [] 列表
    print('{} : {}'.format(letter, m[letter]))
#3 不支持字符串单词
text2 = "hello ,this is a tree ,this ,that..."
m2 = collections.Counter(text2)
for letter in {'this'}:
    print('{} : {}'.format(letter, m2[letter]))

###################输出#################
a : 3
b : 3
c : 1
d : 3
e : 0
this : 1
this : 0

 5，elements() 方法返回一个迭代器，返回所有元素，不能保证元素顺序不变，小于或等于0的元素不包含在内。

import collections
c = collections.Counter('extremely')
print(type(c))
c['z']=0
print(c)
print(type(c.elements()))
print(list(c.elements()))
################输出################
<class 'collections.Counter'>
Counter({'e': 3, 'x': 1, 't': 1, 'r': 1, 'm': 1, 'l': 1, 'y': 1, 'z': 0})
<class 'itertools.chain'>
['e', 'e', 'e', 'x', 't', 'r', 'm', 'l', 'y']

 

6，统计系统字典内所有单词中出现字母，以一个频度分布，然后打印（3条最常见的），如果不向most_common()提供参数，则会生成所有元素构成的一个列表。

# cat /tmp/words 
a
aaa
abc
ace
b
bb
bca
ce
coow
ccc
ddw
wer
#######
# cat collections_counter.py
import collections

c = collections.Counter()
with open('/tmp/words','rt') as f:
    for line in f:
        c.update(line.rstrip().lower())

print('MOst common:')
for letter, count in c.most_common(3):
    print('{}:{:>7}'.format(letter,count))
#######
# cat collections_counter.py
import collections

c = collections.Counter()
with open('/tmp/words','rt') as f:
    for line in f:
        c.update(line.rstrip().lower())

print('MOst common:')
for letter, count in c.most_common():
    print('{}:{:>7}'.format(letter,count))
#################输出############
MOst common:
c:      8
a:      7
b:      5
####
MOst common:
c:      8
a:      7
b:      5
e:      3
w:      3
o:      2
d:      2
r:      1

 

 

 

 

 

 

 

 

 

 

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