从文本fi创建表时，输出会重复多次

Question

使用文本文件中的值创建表时遇到一些问题。我的文本文件如下所示：

e432,6/5/3,6962,c8429,A,4324
e340,2/3/5,566623,c1210,A,3201
e4202,6/5/3,4232,c8419,E,4232
e3230,2/3/5,66632,c1120,A,53204
e4202,6/5/3,61962,c8429,A,4322

我想生成一个包含数组的表，其中：最后一列（amountpaid）值小于第三列（final total），如果第五列（status）等于“a”。未付总额是减去最终总额和已付金额后得出的。你知道吗

生成表的代码是：

data = open("pJoptionc.txt", "r")
info=data.readlines()
data.close
for li in info:
        status=li.split(",")[4]
        finaltotal=int(li.split(",")[2])
        amountpaid=int(li.split(",")[5])
        totalrev=0
        headers = ["Estimate Number", "Date", "Final Total", "Customer Number", "Status", "Amount Paid", "Outstanding Amount"]
        print("    ".join(headers))
        for line in open("pJoptionc.txt", "r"):
            line = line.strip().split(",")
            line.append(str(int(line[2]) - int(line[5])))
            if line[2] == line[5] or line[4] in ("E"):
                continue
            for i, word in enumerate(line):
                print(word.ljust(len(headers[i - (i > 4)])), end="    " * ((i - (i > 4)) != len(headers) - 1))
            print()
            outstandingrev =(finaltotal) - (amountpaid) 
    totalrev += int(outstandingrev)
    print("The total amount of outstanding revenue is...")
    print("£",totalrev)

我想要的输出是

Estimate Number    Date    Final Total    Customer Number    Status    Amount Paid    Outstanding Amount
e432               6/5/3    6962           c8429              A         4324           2638           
e340               2/3/5    566623         c1210              A         3201           563422         
e3230              2/3/5    66632          c1120              A         53204          13428          
e4202              6/5/3    61962          c8429              A         4322           57640          
The total amount of outstanding revenue is...
£ 13428

但是，当我运行代码时，输出是反复重复的表，并且负值出现在“未付金额”列中。我使用的是python3.4.3。你知道吗

Answer 1

该表一遍又一遍地重复，因为您读取了脚本顶部的文件，然后对文件中的每一行重新打开该文件。你知道吗

下面是一种使用namedtuple的方法：

from collections import namedtuple

# define a named tuple for storing each line (record)
fields = ['Estimate_Number', 'Date', 'Final_Total', 'Customer_Number', 'Status', 'Amount_Paid']
record = namedtuple('record', fields)

# first print the header
for field in fields:
    print('%-17s' % field.replace('_', ' '), end='')
print('Outstanding')

totalOutstanding = 0

# open the file using a context manager (the with keyword)
with open("pJoptionc.txt", "r") as data:

    for line in data:

        # strip the newline character, split on the comma and create a record
        row = record(*line.strip().split(','))

        # we can access each field by name now to make the code more readable  
        outstanding = int(row.Final_Total) - int(row.Amount_Paid)

        # print the output
        if outstanding:
            for field in fields:
                print('%-17s' % getattr(row, field), end='')
            print(outstanding)
            totalOutstanding += outstanding

print('Total Outstanding is... %s' % totalOutstanding)

参考文献：

• namedtuple
• string formatting
• argument unpacking

Answer 2

在循环中再次读取该文件，以便它对表中的每一行都重复。尝试这样做： （注意这次文件行上只有1for个循环）

headers = ["Estimate Number", "Date", "Final Total", "Customer Number", "Status", "Amount Paid", "Outstanding Amount"]
print("    ".join(headers))
data = open("pJoptionc.txt", "r")
info=data.readlines()
data.close()
totalrev=0
for li in info:
    line = li.strip().split(",")
    status=line[4]
    finaltotal=int(line[2])
    amountpaid=int(line[5])

    line.append(str(int(line[2]) - int(line[5])))
    if line[2] == line[5] or line[4] in ("E"):
        continue
    for i, word in enumerate(line):
        print(word.ljust(len(headers[i - (i > 4)])), end="    " * ((i - (i > 4)) != len(headers) - 1))
    print()
    outstandingrev =(finaltotal) - (amountpaid) 
    totalrev += int(outstandingrev)
print("The total amount of outstanding revenue is...")
print("£",totalrev)

**旁注：您可能想查看csv模块： https://docs.python.org/3/library/csv.html