Question

我已经问了同样的问题，但我觉得我说的是非常模糊的。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]

进入：

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)

我已经重复了这一点，以表明我可以获得一个包含100名员工的文件，这些文件需要重新排列并按顺序打印出来作为元组。所有项目都具有相同的位置，例如number[0]，salary[1]，position [2]然而，其他名称的中间部分可能有多个名称，因此它没有明确的位置。但是，名字只有一个项目，因此可以找到list1[-1]。

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)

ex1（）

Answer 1

如果输入数据有效，则可以轻松完成重新排序。

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

仅当您的输入值是列表列表时才会起作用。您的样本无效。

Answer 2

我的解决方案是列表清单，但很简单。

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

<强>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，我这样写，所以它很容易理解，因为它是基本的操作。

<强>元组：

如果你想要它是tupels列表只需重写这一行：

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

使用tuple（）从python docs看到解释here。

<强>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

<强>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')]

列表位置更改

2 个答案: