列表位置更改

Question

我已经问过同样的问题了，但是我觉得我是怎么说的很模糊。 基本上我需要改变：

data = 
[['15674' '24000' 'Manager' 'Gregory the 1st' 'John'], 
['15674' '24000' 'Manager' 'Gregory the 1st' 'John'], 
['15674' '24000' 'Manager' 'Gregory the 1st' 'John'],
['15674' '24000' 'Manager' 'Gregory the 1st' 'John']] 

data = [number, salary,position, othernames, firstname] 

分为：

data1= 
('John', 'Gregory the 1st',15674,'Manager',24000),
('John', 'Gregory the 1st',15674,'Manager',24000), 
('John', 'Gregory the 1st',15674,'Manager',24000,) 
('John', 'Gregory the 1st',15674,'Manager',24000)

data1 = (Firstname, othernames, number, position,salary)

我重复了一遍，以表明我可以得到一个包含100名员工的文件，这些员工需要重新安排他们的信息，并按这个顺序以元组的形式打印出来。所有项目都将具有相同的位置，例如number[0]、salary[1]、position [2]，但是中间部分即其他名称可能有多个名称，因此它不会有一个明确的位置。但是名字只能是一个项目，因此可以通过list1[-1]找到。你知道吗

def ex1 ():
    b= input("Please enter a file name to be opened: ")
    a = (b+".txt")    
    data =[]    
def employee(lanme, oname, num,title,salary):
    return (lanme, oname, num, title, salary)

def readfile(a):
    try:
        data =[]
        check = open(a, 'r')
        line =check.readlines()
        for items in line:
            breakup= items.split()
            data.append(breakup)
    except IOError as e :
        print("Failed to open", fileName)
readfile(a)

例1（）

Answer 1

如果您的输入数据是有效的，则可以很容易地完成重新排序。你知道吗

# First, Last, Number, Position, Salary
data = [(f,l,n,p,s) for n,s,p,l,f in data]

只有当您的输入值是一个列表列表时，这才会起作用。你的样品无效。你知道吗

Answer 2

我的解决方案是列表列表，但很简单。你知道吗

data = ["15674 24000 Manager Gregory the 1st John", "15674 24000 Manager Gregory the 1st John",
        "15674 24000 Manager Gregory the 1st John", "15674 24000 Manager Gregory the 1st John"]

data1 = []

for items in data:
    splitNames = items.split()
    number, salary, position, first, des1, des2, last = splitNames
    data1.append([last, first + ' ' + des1 + ' ' + des2, number, position, salary])

for items in data1:
    print items

结果：

['John', 'Gregory the 1st', '15674', 'Manager', '24000']
['John', 'Gregory the 1st', '15674', 'Manager', '24000']
['John', 'Gregory the 1st', '15674', 'Manager', '24000']
['John', 'Gregory the 1st', '15674', 'Manager', '24000']

这不是最好的解决方案，我是这样写的，所以很容易理解，因为这是基本的操作。你知道吗

元组：

如果您想让它成为元组列表，只需重写这一行：

data1.append(tuple([last, first + ' ' + des1 + ' ' + des2, number, position, salary]))

通过使用tuple（），请参阅python文档中的解释here。你知道吗

结果：

('John', 'Gregory the 1st', '15674', 'Manager', '24000')
('John', 'Gregory the 1st', '15674', 'Manager', '24000')
('John', 'Gregory the 1st', '15674', 'Manager', '24000')
('John', 'Gregory the 1st', '15674', 'Manager', '24000')

适应您的代码：

for items in line:
    breakup= items.split()
    number, salary, position, first, des1, des2, last = breakup
    data.append(tuple([last, first + ' ' + des1 + ' ' + des2, number, position, salary]))

print data

结果：

[('John', 'Gregory the 1st', '15674', 'Manager', '24000'), ('John', 'Gregory the 1st', '15674', 'Manager', '24000'), ('John', 'Gregory the 1st', '15674', 'Manager', '24000'), ('John', 'Gregory the 1st', '15674', 'Manager', '24000')]