python3 setdefault的妙用

當字典 d[k]找不到正確的鍵時,Python會拋出異常,有沒有一種優雅的方法來避免這種狀況呢?答案是確定的.

index0.py 從索引中獲取單詞出現的頻率信息,並寫入列表 --沒有使用dict.setdefault

#!/usr/bin/env python
# coding=utf-8
import sys, re

WORD_RE = re.compile(r'\w+')

index = {}
with open(sys.argv[1], encoding='utf-8') as fp:
    for line_no, line in enumerate(fp, 1):
        for match in WORD_RE.finditer(line):
            word = match.group()
            column_no = match.start()+1
            location = (line_no, column_no)
            occurrences = index.get(word, [])
            occurrences.append(location)
            index[word] = occurrences

for word in sorted(index, key=str.upper):
    print(word, index[word])

zen.txt

The Zen of Python, by Tim Peters

Beautiful is better than ugly.
Explicit is better than implicit.
Simple is better than complex.
Complex is better than complicated.
Flat is better than nested.
Sparse is better than dense.
Readability counts.
Special cases aren't special enough to break the rules.
Although practicality beats purity.
Errors should never pass silently.
Unless explicitly silenced.
In the face of ambiguity, refuse the temptation to guess.
There should be one-- and preferably only one --obvious way to do it.
Although that way may not be obvious at first unless you're Dutch.
Now is better than never.
Although never is often better than *right* now.
If the implementation is hard to explain, it's a bad idea.
If the implementation is easy to explain, it may be a good idea.
Namespaces are one honking great idea -- let's do more of those!

執行 python3 index0.py zen.txt

---

a [(19, 48), (20, 53)]
Although [(11, 1), (16, 1), (18, 1)]
ambiguity [(14, 16)]
and [(15, 23)]
are [(21, 12)]
aren [(10, 15)]
at [(16, 38)]
bad [(19, 50)]
be [(15, 14), (16, 27), (20, 50)]
beats [(11, 23)]
Beautiful [(3, 1)]
better [(3, 14), (4, 13), (5, 11), (6, 12), (7, 9), (8, 11), (17, 8), (18, 25)]
break [(10, 40)]
by [(1, 20)]
cases [(10, 9)]
...

index.py 使用了dict.setdefault 只用了一行就解決了獲取和更新單詞的出現狀況列表

#!/usr/bin/env python
# coding=utf-8
import sys, re

WORD_RE = re.compile(r'\w+')

index = {}
with open(sys.argv[1], encoding='utf-8') as fp:
    for line_no, line in enumerate(fp, 1):
        for match in WORD_RE.finditer(line):
            word = match.group()
            column_no = match.start()+1
            location = (line_no, column_no)
            index.setdefault(word, []).append(location)

for word in sorted(index, key=str.upper):
    print(word, index[word])

也就是說:

my_dict.setdefault(key, []).append(new_value)

等價於

if key not in my_dict:
    my_dict[key] = []
my_dict[key].append(new_value)

兩者效果相同,只是setdefault只需一次就完成整個操做,然後者須要進行兩次查詢