Python：迭代问题

Question

我正在对一个文本文件执行文本处理，并尝试迭代到for循环中。你知道吗

fields = [1, 2, 3, 4, 5]

i = 0
with open('file path', 'r') as f:
    for line in f:
        # while i is smaller than the number of fields (=5)
        while i <= len(fields)-1:
            currentfield = fields[i]
            # if the first character of the line matches currentfield
            # (that being a number)
            if line[0] == currentfield:
                print(line[4:])  # print the value in the "third column"
            i += 1

文本文件“f”有如下内容（虚线之间的数字表示年份，每年都有自己的“条目”）：

-------------2000--------------
1        17824
2        20131125192004.9
3        690714s1969    dcu           000 0 eng
4    a       75601809 
4    a    DLC
4    b    eng
4    c    DLC
5    a    WA 750
-------------2001--------------
1        3224
2        20w125192004.9
3        690714s1969    dcu           000 0 eng
5    a    WA 120
-------------2002--------------
1        6563453
2        2013341524626245.9
3        484914s1969    dcu           000 0 eng
4    a       75601809 
4    a    eng
4    c    DLC
5    a    WA 345

文本文件中实际上没有列，但字段编号（即1、2、3、4、5）和后面的值（即17824）之间的空格有两个制表符。我只是不知道怎么打电话给17824。你知道吗

我要做的是迭代每个条目/年份的所有字段，但输出仅给出第一个字段的值1。因此，我得到如下输出：

17824    
3224     
6563453

它只迭代第一个字段，而不是迭代所有字段。如何修复代码，以便将输出创建为类似于表的形式，并在其中迭代字段2、3、4和5？像这样：

17824    20131125192004.9    690714s1969    dcu           000 0 eng  ...and so on
3224     20w125192004.9      690714s1969    dcu           000 0 eng  ...and so on
6563453  2013341524626245.9  484914s1969    dcu           000 0 eng  ...and so on

编辑：我知道我不是很清楚，所以我添加了一些部分。你知道吗

Answer 1

这将帮助您：

for line in f:
    print '\nline[0] is %s' % line[0]
    for currentfield in fields: # loop through all fields
        # convert currentfield to string
        if line[0] == str(currentfield): #if the first character of the line matches currentfield (that being a number)
            print 'Printing field %d' % current field # debugging
            print line[4:] #print the value in the "third column"

这给了我：

u'''line[0] is -

line[0] is 1
Printing field 1
    17824

line[0] is 2
Printing field 2
    20131125192004.9

line[0] is 3
Printing field 3
    690714s1969    dcu           000 0 eng

line[0] is 4
Printing field 4
a      75601809 

line[0] is 4
Printing field 4
a   DLC

line[0] is 4
Printing field 4
b   eng

line[0] is 4
Printing field 4
c   DLC

line[0] is 5
Printing field 5
a   WA 750

line[0] is -

line[0] is 1
Printing field 1
    3224

line[0] is 2
Printing field 2
    20w125192004.9

line[0] is 3
Printing field 3
    690714s1969    dcu           000 0 eng

line[0] is 5
Printing field 5
a   WA 120

line[0] is -

line[0] is 1
Printing field 1
    6563453

line[0] is 2
Printing field 2
    2013341524626245.9

line[0] is 3
Printing field 3
    484914s1969    dcu           000 0 eng

line[0] is 4
Printing field 4
a      75601809 

line[0] is 4
Printing field 4
a   eng

line[0] is 4
Printing field 4
c   DLC

line[0] is 5
Printing field 5
a   WA 345'''

顺便说一下，将line[:4]更改为line[:8]将得到上面粘贴的数据的第三列。你知道吗

然后可以使用regex删除第三列数据后空格后面的任何内容。你知道吗

---

为更改的Q编辑

在这里，我将每一行连接起来，并用l = [el for el in ''.join(line) if el != '']删除列中的所有空格。然后可以通过直接引用列来索引该列，例如对于第4列：l[4]

for line in f:
    l = [el for el in ''.join(line) if el != '']
    print '\nline[0] is %s' % line[0]
    for currentfield in fields: # loop through all fields
        # convert currentfield to string
        if l[0] == str(currentfield): #if the first character of the line matches currentfield (that being a number)
            print 'Printing field %d' % current field # debugging
            print l[currentfield] #print the value in the "third column"