在python中通过ID计算单词的出现次数

时间:2012-04-03 13:23:10

标签: python string count

以下是文件的内容,我的问题是如何计算不同ID的单词“optimus”的出现次数

    ID67    DATEUID Thank you for choosing Optimus prime. Please wait for an Optimus prime to respond. You are currently number 0 in the queue. You should be connected to an agent in approximately TIMEUID.. You are now chatting with AGENTUID   0
    ID67    Optimus MEMORYUID Hi there! Welcome to Optimus prime Web Chat. How can I help you today?        1       
    ID67    Optimus DATEUID I like to pay  prepaid from CURRENCYUID with NUMBERUID expiry on whateve date. my phone no is PHONEUID 2
    ID12120 0 0 0 is the number. They are open 0/0 so you can ring them anytime. SMILEUID   1
    ID12120 Thanks Optimus, I will give them a call. Thanks for your help! HELPUID  2
    ID5552  is the number. They are open 0/0 so you can ring them anytime. SMILEUID 1
    ID5552  Thanks Optimus, I will give them a call. Thanks for your help! HELPUID  2

for line in chat.txt:
   print line, ####print lines and count optimus word for the particular id..

输出应该像

ID67:4
ID12120
ID5552:1

2 个答案:

答案 0 :(得分:1)

一种方法是使用defaultdict作为计数:

from collections import defaultdict
d = defaultdict(int)
with open("chat.txt") as f:
    for line in f:
        id, data = line.split(None, 1)
        d[id] += data.lower().count("optimus")

答案 1 :(得分:1)

>>> from collections import Counter
>>> c = Counter()
>>> for line in chat.txt:
...     c[line.strip().split(" ",1)[0]] += line.count("Optimus")
>>> c
Counter({'ID67': 5, 'ID5552': 1, 'ID12120': 1, '': 0})

您可以将值打印为:

>>> for k, v in c.items():
...     print("{}:{}".format(k, v))
... 
:0
ID67:5
ID5552:1
ID12120:1