我想打印文本文件中存在的唯一行。
例如：如果我的文本文件的内容是：
12345
12345
12474
54675
35949
35949
74564
我想要我的Python程序打印：
12474
54675
74564
我使用的是Python 2.7。
答案 0 :(得分：2)
试试这个：
from collections import OrderedDict
seen = OrderedDict()
for line in open('file.txt'):
line = line.strip()
seen[line] = seen.get(line, 0) + 1
print("\n".join([k for k,v in seen.items() if v == 1]))
打印
12474
54675
74564
更新：感谢下面的评论，这甚至更好：
from collections import Counter, OrderedDict
class OrderedCounter(Counter, OrderedDict):
pass
with open('file.txt') as f:
seen = OrderedCounter([line.strip() for line in f])
print("\n".join([k for k,v in seen.items() if v == 1]))
答案 1 :(得分：2)
使用index()检查列表中每个元素的出现次数，并使用for循环中的with open("file.txt","r")as f:
data=f.readlines()
for x in data:
if data.count(x)>1: #if item is a duplicate
for i in range(data.count(x)):
data.pop(data.index(x)) #find indexes of duplicates, and remove them
with open("file.txt","w")as f:
f.write("".join(data)) #write data back to file as string
删除每个匹配项：
12474
54675
74564
file.txt的：
{{1}}
答案 2 :(得分：2)
您可以使用OrderedDict和Counter删除重复项并维护订单：
from collections import OrderedDict, Counter
class OrderedCounter(Counter, OrderedDict):
pass
with open('/tmp/hello.txt') as f:
ordered_counter = OrderedCounter(f.readlines())
new_list = [k.strip() for k, v in ordered_counter.items() if v==1]
# ['12474', '54675', '74564']
答案 3 :(得分：0)
效率最高，因为它使用count但很简单：
with open("input.txt") as f:
orig = list(f)
filtered = [x for x in orig if orig.count(x)==1]
print("".join(filtered))