1. 读取文件
fin = open("D:\tools\log\words.txt") for fi in fin: print(fi.strip()) #去掉空格
2. 字符串处理
testStr = "asdfjlsj" print(testStr[2:4]) #字符串截取 print(testStr[5:]) print(testStr[6]) print(len(testStr))
print(testStr.isupper())
3. 列表操作(数组)
arr = [1,"323", 2.22, [3,"dsljf"]] arr[1] = 30 print(arr[1]) print(2.22 in arr) for ar in arr: print(type(ar)) print(ar) print(arr[1:3]) #列表拆分 arr.append("hello") #追加数据 print(arr) t1 = [1,2,3] t2 = [4,5] t1.extend(t2) #列表添加 print(t1) print(sum(t1)) t1.pop(1) #通过下标删除元素 print(t1) t1.remove(4) #通过值删除元素 print(t1) s = "we332dfj" print(list(s)) #字符串转列表 sp = "-" arr1 = ["dsf","3r3", "323jj"] s1 = sp.join(arr1) #列表转字符串 print(s1)
4. 字典
eng = dict() #空字典 eng["a"] = "abc" print(eng["a"]) eng2 = {"a":"abc", "b":"bcd"} print(eng2) print(len(eng2)) print("a" in eng2) print(eng2.values()) eng3 = "abcajjkedabedd" eng4 = dict() for st in eng3: if st in eng4: eng4[st] += 1 else: eng4[st] = 1 print(eng4) t1 = "234234@qq.com" name,url = t1.split("@") print(name + " " + url) t = divmod(7, 3) #同时计算出商和余数 也可以使用 quot, rem = divmod(7, 3) 来接收结果 print(t) print(max(t) + min(t)) #获取元祖最大值和最小值 a = "abc" b = [1,2,3] c = zip(a,b) #拉链操作 结果:('a', 1)('b', 2)('c', 3) for d in c: print(d) for index, element in enumerate("asd234"): #enumerate函数遍历下标和值 print("index:"+str(index) + " element:"+element) d = {"a":1, "b":2} #字典转元祖列表 print(items(d)) #运行结果: [('a', 1), ('b', 2)] for key, value in d.items(): print(key + " "+ str(value)) t = [("a",1),("b",2)] print(dict(t)) #元祖的列表转字典 print(random . randint (5 , 10)) #返回一个 5-10 之间的随机数 t2 = [2,3,5] print(random.choice(t2))
5. os 操作文件
#遍历获取目录下的所有文件,包括子文件 def abc(dirName): for name in listdir(dirName): path = os.path.join(dirName, name) #获取全路径(接收一个目录 和 文件名,并把它们合并成一个完整路径) if os.path.isfile(path): print(path) else: abc(path)
print(os.path.exists(dirName)) #路径是否存在 print(os.path.isfile(dirName)) print(os.path.isdir(dirName)) print(os.getcwd()) #获取当前文件所在目录 try: fin = os.open(dirName) except: print("some exception!!!")