仅打印.csv文件中的最后8个条目

Question

我有一个输入的csv文件，如下所示，我只想打印最新的8个条目。.有人可以提供有关如何执行此操作的输入吗？

INPUT:-
trend.csv

['2013-06-25 20:01', '10']
['2013-06-25 20:06', '9']
['2013-06-25 20:06', '8']
['2013-06-26 20:06', '7']
['2013-06-26 20:06', '6']
['2013-06-26 20:06', '5']
['2013-06-26 20:06', '4']
['2013-06-26 20:06', '3']
['2013-06-26 20:06', '2']
['2013-06-26 20:08', '1']

OUTPUT:-
['2013-06-25 20:06', '8']
['2013-06-26 20:06', '7']
['2013-06-26 20:06', '6']
['2013-06-26 20:06', '5']
['2013-06-26 20:06', '4']
['2013-06-26 20:06', '3']
['2013-06-26 20:06', '2']
['2013-06-26 20:08', '1']

码：

import csv
#Now read the recent 8 entries and print
cr = csv.reader(open("trend.csv","rb"))

for row in cr:  
    #print only the recent most 8 entries
    print row

Answer 1

您可以将尾部配方与n = 8的双端队列配合使用。

这将创建一个双头队列，在该队列中，将一个项目添加到末尾（右侧）将有效地弹出一个项目，使它的开头（左侧）不超过最大长度：

>>> from collections import deque
>>> deque(range(10000),8)
deque([9992, 9993, 9994, 9995, 9996, 9997, 9998, 9999], maxlen=8)

csv.reader对象是一个迭代器。 将有限长度的双端队列应用于csv阅读器，您就可以开始了：

import csv
from collections import deque

with open('/tmp/trend.csv','rb') as fin:
    deq=deque(csv.reader(fin),8)

for sub_list in deq:
    print sub_list

以您的10行示例为例，将输出：

['2013-06-25 20:06', '8']
['2013-06-26 20:06', '7']
['2013-06-26 20:06', '6']
['2013-06-26 20:06', '5']
['2013-06-26 20:06', '4']
['2013-06-26 20:06', '3']
['2013-06-26 20:06', '2']
['2013-06-26 20:08', '1']

Answer 2

import csv

# Open the file with a "with" statement to provide automatic cleanup
# in case of exceptions.
with open("trend.csv","rb") as file:
    cr = csv.reader(file)
    lines = [row for row in cr]
# Use slice notation and the wonderful fact that python treats
# negative indices intelligently!
for line in lines[-8:]:
    print line

Answer 3

如果内存/性能不是问题，则可以执行以下操作：

for row in list(cr)[-8:]:  
    print row