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
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)
其他回答
这对我很有效
f = open("history_export.csv", "r")
line= 5
for x in range(line):
a = f.readline()
print(a)
没有特定的方法来读取文件对象暴露的行数。
我想最简单的方法是:
lines =[]
with open(file_name) as f:
lines.extend(f.readline() for i in xrange(N))
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)
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)
Python 3:
with open("datafile") as myfile:
head = [next(myfile) for x in range(N)]
print(head)
Python 2:
with open("datafile") as myfile:
head = [next(myfile) for x in xrange(N)]
print head
下面是另一种方法(Python 2和3都是):
from itertools import islice
with open("datafile") as myfile:
head = list(islice(myfile, N))
print(head)