下面的代码创建了一个函数Which_Line_for_Position(pos),它给出了位置pos的行号,也就是说,位于文件中位置pos的字符所在的行号。
这个函数可以使用任何位置作为参数,独立于文件指针当前位置的值和函数调用之前这个指针移动的历史。
因此,使用此功能,人们不仅限于在行的不间断迭代期间确定当前行的编号,就像 Greg Hewgill 的解决方案一样。
with open(filepath,'rb') as f:
GIVE_NO_FOR_END = {}
end = 0
for i,line in enumerate(f):
end += len(line)
GIVE_NO_FOR_END[end] = i
if line[-1]=='\n':
GIVE_NO_FOR_END[end+1] = i+1
end_positions = GIVE_NO_FOR_END.keys()
end_positions.sort()
def Which_Line_for_Position(pos,
dic = GIVE_NO_FOR_END,
keys = end_positions,
kmax = end_positions[-1]):
return dic[(k for k in keys if pos < k).next()] if pos<kmax else None
.
可以借助模块fileinput编写相同的解决方案:
import fileinput
GIVE_NO_FOR_END = {}
end = 0
for line in fileinput.input(filepath,'rb'):
end += len(line)
GIVE_NO_FOR_END[end] = fileinput.filelineno()
if line[-1]=='\n':
GIVE_NO_FOR_END[end+1] = fileinput.filelineno()+1
fileinput.close()
end_positions = GIVE_NO_FOR_END.keys()
end_positions.sort()
def Which_Line_for_Position(pos,
dic = GIVE_NO_FOR_END,
keys = end_positions,
kmax = end_positions[-1]):
return dic[(k for k in keys if pos < k).next()] if pos<kmax else None
但是这个解决方案有一些不便:
- 它需要导入模块文件输入
- 它会删除文件的所有内容!我的代码中一定有问题,但我不知道fileinput足以找到它。还是fileinput.input()函数的正常行为?
- 似乎在启动任何迭代之前首先完全读取该文件。如果是这样,对于一个非常大的文件,文件的大小可能会超过 RAM 的容量。我不确定这一点:我尝试使用 1,5 GB 的文件进行测试,但它相当长,我暂时放弃了这一点。如果这一点是正确的,它就构成了使用enumerate()的其他解决方案的论据
.
例子:
text = '''Harold Acton (1904–1994)
Gilbert Adair (born 1944)
Helen Adam (1909–1993)
Arthur Henry Adams (1872–1936)
Robert Adamson (1852–1902)
Fleur Adcock (born 1934)
Joseph Addison (1672–1719)
Mark Akenside (1721–1770)
James Alexander Allan (1889–1956)
Leslie Holdsworthy Allen (1879–1964)
William Allingham (1824/28-1889)
Kingsley Amis (1922–1995)
Ethel Anderson (1883–1958)
Bruce Andrews (born 1948)
Maya Angelou (born 1928)
Rae Armantrout (born 1947)
Simon Armitage (born 1963)
Matthew Arnold (1822–1888)
John Ashbery (born 1927)
Thomas Ashe (1836–1889)
Thea Astley (1925–2004)
Edwin Atherstone (1788–1872)'''
#with open('alao.txt','rb') as f:
f = text.splitlines(True)
# argument True in splitlines() makes the newlines kept
GIVE_NO_FOR_END = {}
end = 0
for i,line in enumerate(f):
end += len(line)
GIVE_NO_FOR_END[end] = i
if line[-1]=='\n':
GIVE_NO_FOR_END[end+1] = i+1
end_positions = GIVE_NO_FOR_END.keys()
end_positions.sort()
print '\n'.join('line %-3s ending at position %s' % (str(GIVE_NO_FOR_END[end]),str(end))
for end in end_positions)
def Which_Line_for_Position(pos,
dic = GIVE_NO_FOR_END,
keys = end_positions,
kmax = end_positions[-1]):
return dic[(k for k in keys if pos < k).next()] if pos<kmax else None
print
for x in (2,450,320,104,105,599,600):
print 'pos=%-6s line %s' % (x,Which_Line_for_Position(x))
结果
line 0 ending at position 25
line 1 ending at position 51
line 2 ending at position 74
line 3 ending at position 105
line 4 ending at position 132
line 5 ending at position 157
line 6 ending at position 184
line 7 ending at position 210
line 8 ending at position 244
line 9 ending at position 281
line 10 ending at position 314
line 11 ending at position 340
line 12 ending at position 367
line 13 ending at position 393
line 14 ending at position 418
line 15 ending at position 445
line 16 ending at position 472
line 17 ending at position 499
line 18 ending at position 524
line 19 ending at position 548
line 20 ending at position 572
line 21 ending at position 600
pos=2 line 0
pos=450 line 16
pos=320 line 11
pos=104 line 3
pos=105 line 4
pos=599 line 21
pos=600 line None
.
然后,有了函数Which_Line_for_Position(),很容易获得当前行的编号:只需将f.tell()作为参数传递给函数
但是警告:当使用f.tell()并在文件中移动文件指针时,绝对有必要以二进制模式打开文件:'rb' or 'rb+' or 'ab' or ....