Python每日一練0010

jayzou 發(fā)布于2019-07-30 16:02 / 1614人閱讀

摘要：問題你有一個字典或者實例的序列，然后你想根據(jù)某個特定的字段來分組迭代訪問。

問題

你有一個字典或者實例的序列，然后你想根據(jù)某個特定的字段來分組迭代訪問。

解決方案

使用itertools.groupby() 函數(shù)

假設(shè)有下列的字典列表：

rows = [
    {"address": "5412 N CLARK", "date": "07/01/2012"},
    {"address": "5148 N CLARK", "date": "07/04/2012"},
    {"address": "5800 E 58TH", "date": "07/02/2012"},
    {"address": "2122 N CLARK", "date": "07/03/2012"},
    {"address": "5645 N RAVENSWOOD", "date": "07/02/2012"},
    {"address": "1060 W ADDISON", "date": "07/02/2012"},
    {"address": "4801 N BROADWAY", "date": "07/01/2012"},
    {"address": "1039 W GRANVILLE", "date": "07/04/2012"},
]

現(xiàn)在按照date字段來分組訪問，就可以使用itertools.groupby()了

from itertools import groupby
from operator import itemgetter
rows.sort(key=itemgetter("date"))
for key, group in groupby(rows, key=itemgetter("date")):
    print(key)
    for item in group:
        print(4 * " ", item)

輸出

07/01/2012
     {"address": "5412 N CLARK", "date": "07/01/2012"}
     {"address": "4801 N BROADWAY", "date": "07/01/2012"}
07/02/2012
     {"address": "5800 E 58TH", "date": "07/02/2012"}
     {"address": "5645 N RAVENSWOOD", "date": "07/02/2012"}
     {"address": "1060 W ADDISON", "date": "07/02/2012"}
07/03/2012
     {"address": "2122 N CLARK", "date": "07/03/2012"}
07/04/2012
     {"address": "5148 N CLARK", "date": "07/04/2012"}
     {"address": "1039 W GRANVILLE", "date": "07/04/2012"}

討論

itertools.groupby迭代器和一個可選的key參數(shù)，按key來分組的，如果key是None的話，則按元素分組

itertools.groupby返回每個不同的key和一個迭代器對象，這個迭代器對象就是key對應(yīng)的一組元素

并且itertools.groupby要求分組之前，迭代器的所有元素必須是有序的，原因跟itertools.groupby的實現(xiàn)有關(guān)

itertools.groupby()大致實現(xiàn)：

class groupby:
    # [k for k, g in groupby("AAAABBBCCDAABBB")] --> A B C D A B
    # [list(g) for k, g in groupby("AAAABBBCCD")] --> AAAA BBB CC D
    def __init__(self, iterable, key=None):
        if key is None:
            key = lambda x: x
        self.keyfunc = key
        self.it = iter(iterable)
        self.tgtkey = self.currkey = self.currvalue = object()
    def __iter__(self):
        return self
    def __next__(self):
        while self.currkey == self.tgtkey:
            self.currvalue = next(self.it)    # Exit on StopIteration
            self.currkey = self.keyfunc(self.currvalue)
        self.tgtkey = self.currkey
        return (self.currkey, self._grouper(self.tgtkey))
    def _grouper(self, tgtkey):
        while self.currkey == tgtkey:
            yield self.currvalue
            try:
                self.currvalue = next(self.it)
            except StopIteration:
                return
            self.currkey = self.keyfunc(self.currvalue)

來源

Python Cookbook

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