隔离动态结构化数据

问题描述 投票:1回答:1

我有大量数据(大约20K行),如下所示。

Caller1 5:30AM Mexico USA 2-22-19
Caller2 1:30AM Mexico USA 2-22-19
Caller3 2:30AM Mexico USA 2-22-19
Caller1 5:30AM Mexico USA 2-22-19
Caller5 3:30AM Mexico USA 2-22-19
Caller3 4:30AM Mexico USA 2-22-19
Caller2 5:30AM Mexico USA 2-22-19
Caller1 7:30AM Mexico USA 2-22-19
Caller12 9:39AM Mexico USA 2-22-19
Caller14 8:36AM Mexico USA 2-22-19
Caller15 2:39AM Mexico USA 2-22-19
Caller16 3:32AM Mexico USA 2-22-19

我正在寻找一种基于CallerID隔离数据的方法,如下所示:

Caller1 5:30AM Mexico USA 2-22-19
Caller1 5:30AM Mexico USA 2-22-19
Caller1 7:30AM Mexico USA 2-22-19
---------------------------------
Caller2 1:30AM Mexico USA 2-22-19
Caller2 5:30AM Mexico USA 2-22-1
---------------------------------
.
.

我最初习惯将这些数据存储为dictionary,并且任何新数据都被添加到该字典中

我有麻烦隔离,因为初始参数CallerID也是可变的。

我的代码:

>>> input = [('caller1', 'data....'),('caller2','data,,,,,)
>>> from collections import defaultdict
>>> res = defaultdict(list)
>>> for v, k in input: res[k].append(v)

由于数据集太大,我不能使用它

Python中是否有任何软件包可以根据句子的第一个单词来分隔数据?

python sorting machine-learning
1个回答
0
投票

您可以尝试这种方法,将数据存储在列表的字典中,并将键作为要分组的字符串,即Caller1,Caller2等。

     data = ["Caller1 5:30AM Mexico USA 2-22-19",
            "Caller2 1:30AM Mexico USA 2-22-19",
            "Caller3 2:30AM Mexico USA 2-22-19",
            "Caller1 5:30AM Mexico USA 2-22-19",
            "Caller5 3:30AM Mexico USA 2-22-19",
            "Caller3 4:30AM Mexico USA 2-22-19",
            "Caller2 5:30AM Mexico USA 2-22-19",
            "Caller1 7:30AM Mexico USA 2-22-19",
            "Caller12 9:39AM Mexico USA 2-22-19",
            "Caller14 8:36AM Mexico USA 2-22-19",
            "Caller15 2:39AM Mexico USA 2-22-19",
            "Caller16 3:32AM Mexico USA 2-22-19"]

    grouped_data = {}

    # ITERATE THE INPUT AND STORE DATA WITH KEY IN DICTIONARY OF LIST 
    for x in data:
        temp: list = []
        key = x.split(' ')[0]
        if key in grouped_data:
            temp = grouped_data.get(key)
        temp.append(x)
        grouped_data[key] = temp

    # PRINT THE DATA AS GROUPED
    for k, v in grouped_data.items():
        print(f"data for {k}")
        for d in v:
            print(d)
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