隔离动态结构化数据

Question

我有大量的数据(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
---------------------------------
.
.

我最初将此数据存储为dictionary，并将任何新数据添加到该字典中。

由于初始参数CallerID也是可变的，所以我在隔离时遇到了问题。

我的守则：

>>> 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中是否有基于句子的第一个单词来隔离数据的包？

Answer 1

您可以尝试这种方法，将数据存储在带有键的列表字典中，作为要与ie Caller1、Caller2等进行分组的字符串。

     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)