我们有一个很大的原始数据文件,我们想把它修剪成指定的大小。
如何在python中获取文本文件的前N行?所使用的操作系统对实现有任何影响吗?
我们有一个很大的原始数据文件,我们想把它修剪成指定的大小。
如何在python中获取文本文件的前N行?所使用的操作系统对实现有任何影响吗?
当前回答
fname = input("Enter file name: ")
num_lines = 0
with open(fname, 'r') as f: #lines count
for line in f:
num_lines += 1
num_lines_input = int (input("Enter line numbers: "))
if num_lines_input <= num_lines:
f = open(fname, "r")
for x in range(num_lines_input):
a = f.readline()
print(a)
else:
f = open(fname, "r")
for x in range(num_lines_input):
a = f.readline()
print(a)
print("Don't have", num_lines_input, " lines print as much as you can")
print("Total lines in the text",num_lines)
其他回答
对于前5行,简单地做:
N=5
with open("data_file", "r") as file:
for i in range(N):
print file.next()
基于gnibbler的投票结果(2009年11月20日0:27):这个类将head()和tail()方法添加到文件对象。
class File(file):
def head(self, lines_2find=1):
self.seek(0) #Rewind file
return [self.next() for x in xrange(lines_2find)]
def tail(self, lines_2find=1):
self.seek(0, 2) #go to end of file
bytes_in_file = self.tell()
lines_found, total_bytes_scanned = 0, 0
while (lines_2find+1 > lines_found and
bytes_in_file > total_bytes_scanned):
byte_block = min(1024, bytes_in_file-total_bytes_scanned)
self.seek(-(byte_block+total_bytes_scanned), 2)
total_bytes_scanned += byte_block
lines_found += self.read(1024).count('\n')
self.seek(-total_bytes_scanned, 2)
line_list = list(self.readlines())
return line_list[-lines_2find:]
用法:
f = File('path/to/file', 'r')
f.head(3)
f.tail(3)
N = 10
with open("file.txt", "a") as file: # the a opens it in append mode
for i in range(N):
line = next(file).strip()
print(line)
从Python 2.6开始,您可以利用IO基类中更复杂的函数。所以上面评分最高的答案可以改写为:
with open("datafile") as myfile:
head = myfile.readlines(N)
print head
(你不必担心你的文件少于N行,因为没有StopIteration异常抛出。)
有一个简单的方法来获取前10行:
with open('fileName.txt', mode = 'r') as file:
list = [line.rstrip('\n') for line in file][:10]
print(list)