Question

刚刚开始学习Python，我正在努力解决这个问题。

我正在打开一个长度可变的txt文件，我需要一次迭代用户可定义的行数。当我到达文件的末尾时，我在主题字段中收到错误。我还尝试了readlines（）函数和导致问题的“if”语句的几个变体。我似乎无法获得代码来找到EOF。

嗯，当我写这篇文章时，我在想......我是否需要将数据库“EOF”添加到数组中并且只是寻找它？这是找到自定义EOF的最佳解决方案吗？

我的代码段如下：

### variables defined outside of scapy PacketHandler ##
x = 0
B = 0
##########

with open('dict.txt') as f:
    lines = list(f)
    global x
    global B
    B = B + int(sys.argv[3])
        while x <= B:
           while y <= int(sys.argv[2]): 
               if lines[x] != "":
                   #...do stuff...
                   # Scapy send packet Dot11Elt(ID="SSID",info"%s" %               (lines[x].strip()) 
                   # ....more code...
           x = x 1

Answer 1

尝试for循环。您已经创建了列表，现在可以遍历它。

with open('dict.txt') as f:
    lines = list(f)
    for item in lines: #each item here is an item in the list you created
        print(item)

通过这种方式，您可以浏览文本文件的每一行，而不必担心它的结束位置。

编辑：

你也可以这样做！

with open('dict.txt') as f:
    for row in f:
        print(row)

Answer 2

假设您需要一次读取X行，将其放入列表并进行处理：

with open('dict.txt') as f:

    enoughLines = True  

    while enoughLines:
        lines = []

        for i in range(X):
            l = f.readline()
            if l != '':
                lines.append( l )
            else:
                enoughLines = False
                break

        if enoughLines:
            #Do what has to be done with the list “lines”
        else:
            break

#Do what needs to be done with the list “lines” that has less than X lines in it

Answer 3

以下函数将返回一个生成器，该生成器返回文件中的下n行：

def iter_n(obj, n):
    iterator = iter(obj)
    while True:
        result = []
        try:
            while len(result) < n:
                result.append(next(iterator))
        except StopIteration:
            if len(result) == 0:
                raise
        yield result

以下是如何使用它：

>>> with open('test.txt') as f:
...     for three_lines in iter_n(f, 3): 
...         print three_lines
...
['first line\n', 'second line\n', 'third line\n']
['fourth line\n', 'fifth line\n', 'sixth line\n']
['seventh line\n']

test.txt的内容：

first line
second line
third line
fourth line
fifth line
sixth line
seventh line

请注意，因为文件没有3行的倍数，所返回的最后一个值不是3行，而是文件的其余部分。

因为此解决方案使用生成器，所以不要求将完整文件读入内存（放入列表中），而是根据需要对其进行迭代。

实际上，上面的函数可以迭代任何可迭代对象，如列表，字符串等：

>>> for three_numbers in iter_n([1, 2, 3, 4, 5, 6, 7], 3): 
...     print three_numbers
... 
[1, 2, 3]
[4, 5, 6]
[7]
>>> for three_chars in iter_n("1234567", 3): 
...     print three_chars
... 
['1', '2', '3']
['4', '5', '6']
['7']

Answer 4

如果你想在列表中获得n行，请使用itertools.islice产生每个列表：

from itertools import islice
def yield_lists(f,n):
    with open(f) as f:
        for sli in iter(lambda : list(islice(f,n)),[]):
            yield sli

如果你想使用循环，你根本不需要一个while循环，你可以使用范围n-1中的内循环调用文件对象的下一个，默认值为空字符串，如果我们得到的话一个空字符串打破循环，如果不只是追加并再次产生每个列表：

def yield_lists(f,n):
    with open(f) as f:
        for line in f:
            temp = [line]
            for i in range(n-1):
                line = next(f,"")
                if not line:
                    break
                temp.append(line)
            yield temp

读取文件时“IndexError：列表索引超出范围”

4 个答案: