1、
{i:5 for i in [1,2,3] }
执行结果{1: 5, 2: 5, 3: 5}
2、常用指令
1）?显示变量的通用信息
?? 如果可以的话显示源码
a = [1,2,3] a?
np.*load*?
执行结果：
np.__loader__ np.load np.loads np.loadtxt
2）%命令
%run aa.py #执行py文件 %load aa.py #导入py文件 %paste #执行剪切板的代码 %cpaste #执行 %timeit a= [n for n in range(1000)] #计时 %matplotlib inline #显示图片
3) 获取类型
type(5) isinstance(5.0,(int,float)) getattr(‘data‘,‘lower‘) #通过属性获取方法名称 def isiterable(obj): #是否枚举类型 try: iter(obj) return True except TypeError: # not iterable return False isiterable(‘a string‘) 1 is not 2 #检查是否是同样对象) int(‘3‘) #类型转化
type(None)
datetime.strptime(‘20091031 14:59:36‘, ‘%Y%m%d %H:%M:%S‘) #字符串转化为日期
4)表达式
value = true-expr if condition else false-expr
x=5 ‘Non-negative‘ if x >= 0 else ‘Negative‘
5)元组列表
values=1,2,3,4,5 a,b,*_ = values #元组拆包 list((1,2,3)) #元组转列表 alist =[2,3,7,2] 2 in alist #判断是否在列表里 alist + [‘a‘,‘b‘,‘c‘,4] #合并列表
#切片 seq=[7,2,3,7,5,6,0,1] seq[-4:] #[5, 6, 0, 1] 从结尾倒数第四个开始取数 seq[-6:-2] #[6, 3, 5, 6] seq[::2] #[7, 3, 3, 6, 1] 间隔2个取一次数 seq[::-1] #[1,0,6,5,3,6,3,2,7] 反转 some_list = [‘foo‘, ‘bar‘, ‘baz‘] mapping = {} for i, v in enumerate(some_list): mapping[i]=v #{0: ‘foo‘, 1: ‘bar‘, 2: ‘baz‘} 获取值 #zip 生成元组 seq1 = [‘foo‘, ‘bar‘, ‘baz‘] seq2 = [‘one‘, ‘two‘, ‘three‘] zipped = zip(seq1, seq2) list(zipped) #[(‘foo‘, ‘one‘), (‘bar‘, ‘two‘), (‘baz‘, ‘three‘)] for i, (a, b) in enumerate(zip(seq1, seq2)): mapping[i] =(a, b) mapping #{0: (‘foo‘, ‘one‘), 1: (‘bar‘, ‘two‘), 2: (‘baz‘, ‘three‘)} tupt =[(‘foo‘, ‘one‘), (‘bar‘, ‘two‘), (‘baz‘, ‘three‘)] a,b =zip( *tupt ) a,b #((‘foo‘, ‘bar‘, ‘baz‘), (‘one‘, ‘two‘, ‘three‘)) list(reversed(range(10))) #反转 #[9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
6）字典
#序列转字典1 #mapping = {} #for key, value in zip(key_list, value_list): #mapping[key] = value #序列转字典2 mapping = dict(zip(range(5), reversed(range(5)))) mapping #{0: 4, 1: 3, 2: 2, 3: 1, 4: 0} #获取设置字典默认值 value = some_dict.get(key, default_value) words = [‘apple‘, ‘bat‘, ‘bar‘, ‘atom‘, ‘book‘] #方法一 by_letter = {} for word in words: letter=word[0] if letter in by_letter: by_letter[letter].append(word) else: by_letter[letter]= [word] #方法二 for word in words: letter = word[0] by_letter.setdefault(letter, []).append(word) #检查是否可以hash hash(‘string‘) #如果希望数组作为key的话需要转我元组 d={} d[tuple([1, 2, 3])] = 5
7）集合
#创建集合set的2种方式 set([2, 2, 2, 1, 3, 3]) {2,2,2,1,3,3} #输出{1,2,3}
8）list、set、dic
#[expr for val in collection if condition] strings = [‘a‘, ‘as‘, ‘bat‘, ‘car‘, ‘dove‘, ‘python‘] [x.upper() for x in strings if len(x) > 2] #[‘BAT‘, ‘CAR‘, ‘DOVE‘, ‘PYTHON‘] #字典 #dict_comp = {key-expr : value-expr for value in collection if condition} #集合 #set_comp = {expr for value in collection if condition} {len(x) for x in strings} set(map(len, strings)) # 输出：{1, 2, 3, 4, 6} {val : index for index, val in enumerate(strings)} #{‘a‘: 0, ‘as‘: 1, ‘bat‘: 2, ‘car‘: 3, ‘dove‘: 4, ‘python‘: 5} all_data = [[‘John‘, ‘Emily‘, ‘Michael‘, ‘Mary‘, ‘Steven‘],[‘Maria‘, ‘Juan‘, ‘Javier‘, ‘Natalia‘, ‘Pilar‘]] names_of_interest = [] for names in all_data: enough_es = [name for name in names if name.count(‘e‘) >= 2] names_of_interest.extend(enough_es) [name for names in all_data for name in names if name.count(‘e‘) >= 2] #[‘Steven‘]
9) 函数
#yield使用 def squares(n=10): print(‘Generating squares from 1 to {0}‘.format(n ** 2)) for i in range(1, n + 1): yield i**2 gen = squares() for x in gen: print(x,end=‘ ‘) #Generating squares from 1 to 100 #1 4 9 16 25 36 49 64 81 100
原文：https://www.cnblogs.com/excellence/p/12811723.html