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Der Inhalt dieses Artikels befasst sich mit der Verwendung von setdefault in Python3. Er hat einen gewissen Referenzwert. Ich hoffe, dass er für Sie hilfreich ist.
Wenn das Wörterbuch d[k] den richtigen Schlüssel nicht finden kann, löst Python eine Ausnahme aus. Gibt es eine elegante Möglichkeit, diese Situation zu vermeiden?
index0.py Ruft ab Worthäufigkeitsinformationen aus dem Index und schreibt sie in die Liste – ohne 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])
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!
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 verwendet dict.setdefault, um das Problem des Abrufens und Aktualisierens der Vorkommensliste von Wörtern in nur einer Zeile zu lösen
#!/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])
my_dict.setdefault(key, []).append(new_value)
if key not in my_dict: my_dict[key] = [] my_dict[key].append(new_value)
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