Question

我在文件中有现有数据，如下所示：

d893ecee58ee4d6f1ca56a358d2e6287
69
ae0d10efd7663c734b9ea66cec5aaa44
100
c9136ba49f4b1a8e89d6ed35cac95f7c
100
67c1431d8a06d7b2e31g86874b757eeb
0
8478b9587875f65d5afe54f541bada61
11

我想要做的是在文档中搜索数值大于30的任何行，并在该数字上方打印该行。

这就是我现在所拥有的：

with open ('somefile.txt','r') as f, open('newfile.txt','w') as fnew:
        for i, line in enumerate(f):
            if line.startswith('1' or '2' or '3' or '4' or '5' or '6' or '7' or '8' or '9' or '10' or '11' or '12' or '13' or '14' or '15' or '16' or '17' or '18' or '19' or '20' or '21' or '22' or '23' or '24' or '25' or '26' or '27' or '28' or '29' or '30'):
                fnew.write(line -1)

我知道这不是最干净的剧本，但我只想要一些有用的东西。

Answer 1

这是我采取的方法，它假定您的输入数据是常规的：

with open('data.txt') as f:
    while True:
        try:
            data = next(f).strip()
            number = next(f).strip()
        except StopIteration:
            # EOF
            break
        number = int(number)
        if number > 30:
            # TODO: Write data to other file
            print(data)

Answer 2

如果你想要不同的方法，那么在dict中收集数据和值，然后如果key大于30则迭代dict然后获取值：

values={}
with open('test.txt','r') as f:
    sub = []
    for line in f:

        sub.append(line.strip())
        if len(sub) == 2:
            if int(sub[1]) not in values:
                values[int(sub[1])]=[sub[0]]
            else:
                values[int(sub[1])].append(sub[0])

            sub=[]




for key,value in values.items():
    if key>30:
        print(key,value)

输出：

100 ['ae0d10efd7663c734b9ea66cec5aaa44', 'c9136ba49f4b1a8e89d6ed35cac95f7c']
69 ['d893ecee58ee4d6f1ca56a358d2e6287']

一步一步：

首先收集dict中的所有值和数字：

values={}
with open('test.txt','r') as f:
    sub = []
    for line in f:

        sub.append(line.strip())
        if len(sub) == 2:
            if int(sub[1]) not in values:
                values[int(sub[1])]=[sub[0]]
            else:
                values[int(sub[1])].append(sub[0])

            sub=[]

它会给出：

{0: ['67c1431d8a06d7b2e31g86874b757eeb'], 11: ['8478b9587875f65d5afe54f541bada61'], 100: ['ae0d10efd7663c734b9ea66cec5aaa44', 'c9136ba49f4b1a8e89d6ed35cac95f7c'], 69: ['d893ecee58ee4d6f1ca56a358d2e6287']}

现在迭代此dict并获取该键的值，如果键> 30

Answer 3

您可以使用列表推导和itertools.compress的组合来解决此问题。

您的文件格式必须严格遵守您在此处发布的内容。

\0

https://docs.python.org/3/library/itertools.html#itertools.compress

Compress使用2个列表并在list1上返回一个迭代器，它只包含在另一个列表中有import itertools # zero-line is the ID, 1st line is the number. NO empty lines in between. text = '''d893ecee58ee4d6f1ca56a358d2e6287 69 ae0d10efd7663c734b9ea66cec5aaa44 100 c9136ba49f4b1a8e89d6ed35cac95f7c 100 67c1431d8a06d7b2e31g86874b757eeb 0 8478b9587875f65d5afe54f541bada61 11 ''' lines = text.split("\n") # list of all lines - you can get that from # file with readlines() data = lines[0::2] # your data is in every 2nd line starting at 0 # your numbers are in every 2nd line starting on 1 nums = [1 if (int(x) > 30) else 0 for x in lines[1::2] ] # the list comprehension creates a list of 0 and 1 - 1 if number > 30 # itertools.compress does the lifing for you result = itertools.compress(data,nums) print(list(result))的元素。

没有itertools：

True

Answer 4

您的方法存在一些问题。这是我的尝试：

def line_gt_30(line):
   try:
       return int(line.strip()) > 30
   except ValueError:
       return False

prev_line = None
with open ('somefile.txt','r') as f, open('newfile.txt','w') as fnew:
    for line in f:
        if line_gt_30(line) and prev_line is not None: 
            fnew.write(prev_line)
        prev_line = line

我还没有真正测试过，但它应该可行。它类似于你的，但它解决了很多你的问题。

首先，你不需要在这里使用枚举。我不确定你使用print得到了什么（第-1行），但你需要将前一行保留在temp变量中，因为我们的文件现在是一个可迭代的，这意味着你不能使用索引来抓取像一个列表。

同样在开头（'1'或'2'或'3'或'4'或......）相当于startswith（'1'）。你应该仔细阅读或做什么。

Answer 5

你可以试试这个：

data = [i.strip('\n') for i in open('filename.txt')]
final_data = [data[i-1] for i in range(len(data)) if all(b.isdigit() for b in data[i]) and int(data[i]) > 30]

输出：

['d893ecee58ee4d6f1ca56a358d2e6287', 'ae0d10efd7663c734b9ea66cec5aaa44', 'c9136ba49f4b1a8e89d6ed35cac95f7c']

在文档中搜索数字并打印上面的行

5 个答案: