python 一些小 tip

列表去重

以前经常用遍历或者科学计算库等方法，其实有一个很简单的技巧，那就是利用集合的互异性

>>> a
[0, 1, 2, 3, 1]
>>> b = set(a)
>>> b
{0, 1, 2, 3}
>>> list(b)
[0, 1, 2, 3]

列表中去除重复字典

对嵌套的字典无效
[dict(t) for t in set([tuple(d.items()) for d in the_list])]

li = [dict(t) for t in set([tuple(d.items()) for d in li])]

修改删除列表部分值

>>> a = [1, 2, 3, 4, 5]
>>> a[2:3] = [0, 0]
>>> a
[1, 2, 0, 0, 4, 5]
>>> a[1:1] = [8, 9]
>>> a
[1, 8, 9, 2, 0, 0, 4, 5]
>>> a[1:-1] = []
>>> a
[1, 5]

压缩和解压缩

# 压缩
import zipfile
import os
def zipDir(dirpath,outFullName):
    """
    压缩指定文件夹
    :param dirpath: 目标文件夹路径
    :param outFullName: 压缩文件保存路径+xxxx.zip
    :return: 无
    """
    zip = zipfile.ZipFile(outFullName,"w",zipfile.ZIP_DEFLATED)
    for path,dirnames,filenames in os.walk(dirpath):
        # 去掉目标跟路径，只对目标文件夹下边的文件及文件夹进行压缩
        fpath = path.replace(dirpath,'')
        for filename in filenames:
            zip.write(os.path.join(path,filename),os.path.join(fpath,filename))
    zip.close()
print(1)
startdir = "MathAmino/train_models"  #要压缩的文件夹路径
file_news = 'mytrain.zip' # 压缩后文件夹的名字 
zipDir(startdir,file_news)
# 解压缩
# 压缩文件
import zipfile
f = zipfile.ZipFile("/home/kesci/MathAmino.zip",'r')
for file in f.namelist():
    f.extract(file,"/home/kesci/work/")

列表组合为字典

>>> a = [1, 2, 3]
>>> b = ['a', 'b', 'c']
>>> z = zip(a, b)
>>> z
[(1, 'a'), (2, 'b'), (3, 'c')]
>>> zip(*z)
[(1, 2, 3), ('a', 'b', 'c')]

两个一一对应列表转化为字典

>>> a = [1, 2, 3]
>>> b = ['a', 'b', 'c']
>>> z = zip(a, b)
>>> z
[(1, 'a'), (2, 'b'), (3, 'c')]
>>> d = dict(z)
>>> d
{1: 'a', 2: 'b', 3: 'c'}

翻转字典

>>> m = {'a': 1, 'b': 2, 'c': 3, 'd': 4}
>>> m.items()
[('a', 1), ('c', 3), ('b', 2), ('d', 4)]
>>> zip(m.values(), m.keys())
[(1, 'a'), (3, 'c'), (2, 'b'), (4, 'd')]
>>> mi = dict(zip(m.values(), m.keys()))
>>> mi
{1: 'a', 2: 'b', 3: 'c', 4: 'd'}

集合操作

>>> A = {1, 2, 3, 3}
>>> A
set([1, 2, 3])
>>> B = {3, 4, 5, 6, 7}
>>> B
set([3, 4, 5, 6, 7])
>>> A | B
set([1, 2, 3, 4, 5, 6, 7])
>>> A & B
set([3])
>>> A - B
set([1, 2])
>>> B - A
set([4, 5, 6, 7])
>>> A ^ B
set([1, 2, 4, 5, 6, 7])
>>> (A ^ B) == ((A - B) | (B - A))
True

在字符串中找指定字符串位置

使用list的index方法可以找到list中第一次出现该元素的位置
>>> l ``= [``'a'``,``'b'``,``'c'``,``'c'``,``'d'``,``'c'``]
>>> find``=``'b'
>>> l.index(find)
1

找出出现该元素的所有位置可以使用一个简单的表理解来实现
>>> find ``= 'c'
>>> [i ``for i,v ``in enumerate``(l) ``if v``=``=``find]
[``2``, ``3``, ``5``]

列表倒序

>>> x = [1,5,2,3,4]
>>> x.reverse()
>>> x
[4, 3, 2, 5, 1]

找出列表中最大或最小的三个

# -*- coding: utf-8 -*-
import heapq
nums = [1, 8, 2, 23, 7, -4, 18, 23, 24, 37, 2]
# 最大的3个数的索引
max_num_index_list = map(nums.index, heapq.nlargest(3, nums))
# 最小的3个数的索引
min_num_index_list = map(nums.index, heapq.nsmallest(3, nums))
print(list(max_num_index_list))
print(list(min_num_index_list))

列表平均分为n份

# listTemp 为列表 分成每份n个元素的列表
def func(listTemp, n):
    for i in range(0, len(listTemp), n):
        yield listTemp[i:i + n]
平均分为n份
def average_func(m, n):
    f = False
    s = len(m) // n
    lef = len(m) % n
    lop = 0
    stopat = 0
    if lef != 0:
        s += 1
        f = True
    ret = []
    if f:
        for i in range(lef):
            ret.append(m[i*s:(i+1)*s])
            stopat = i*s+1
            lop = i
        s -= 1
        for i in range(1, n-lop):
            ret.append(m[stopat+i*s:stopat+(i+1)*s])
        return ret
    else:
        for i in range(n):
            ret.append(m[i*s:(i+1)*s])
        return ret

统计列表中重复元素的个数

>>> from collections import Counter
>>> Counter([1,2,2,2,2,3,3,3,4,4,4,4])
Counter({2: 4, 4: 4, 3: 3, 1: 1})

查看当前文件夹中文件

for (dirpath, dirnames, filenames) in os.walk("try_test"):
       print(dirpath, dirnames, filenames)

存格式化的json文件

json_str = json.dumps(your_list, ensure_ascii=False, indent=4) # 缩进4字符
with open(your_path, 'w') as json_file:
    json_file.write(json_str)

读取json文件

with open("../config/record.json",'r') as load_f:
    load_dict = json.load(load_f)