Python3：計算串列中的出現次數并將其排序為后代

我有一個有效的代碼，但是我想學習一個更 Pythonic 的版本，因為我覺得我的代碼很笨拙。

想法是計算串列（“名稱”）中出現的次數并將其排序。

aux_list = []
names = ["Jessica", "John", "Steve", "John", "John", "Steve"]
total = {}

for name in names:
    
    if name is not aux_list:
        aux_list.append(name)
        
    count = names.count(name)
    key = { name : count }
    total.update(key)

# sort desc     
sorted_total = sorted(total, key=total.get, reverse=True)

# Desired output: <NAME> <TOTAL> DESC ORDER
# John 3
# Steve 2
# Jessica 1

for r in sorted_total:
    print(r, total[r])

uj5u.com熱心網友回復：

使用collections.Counter：

from collections import Counter

names = ["Jessica", "John", "Steve", "John", "John", "Steve"]
res = Counter(names).most_common()
print(res)

輸出

[('John', 3), ('Steve', 2), ('Jessica', 1)]

uj5u.com熱心網友回復：

以下是一些建議，按與您的代碼的相似性順序列出。第一個直接改編自您的代碼，并稍作修改；最后一個是我會怎么做，與 Dani Mesejo 的回答相同。這些提議利用：

• dict索引total[name]；
• dict方法total.get(name, default=0);
• dict方法total.setdefault(name, default=0);
• dict子類defaultdict；
• dict子類Counter。

# SIMPLE LOOP WITH .COUNT
names = ["Jessica", "John", "Steve", "John", "John", "Steve"]
total = {}
    
for name in names:
    total[name] = names.count(name)
    
sorted_total = sorted(total, key=total.get, reverse=True)
    
for r in sorted_total:
    print(r, total[r])

# SIMPLE LOOP WITH .COUNT, OPTIMIZED
names = ["Jessica", "John", "Steve", "John", "John", "Steve"]
total = {}
    
for name in names:
    if name not in total:     # adding this if avoids recomputing the expensive `.count`
        total[name] = names.count(name)
    
sorted_total = sorted(total, key=total.get, reverse=True)
    
for r in sorted_total:
    print(r, total[r])

# DICT COMPREHENSION WITH .COUNT

names = ["Jessica", "John", "Steve", "John", "John", "Steve"]
total = {name: names.count(name) for name in names}
    
sorted_total = sorted(total, key=total.get, reverse=True)
    
for r in sorted_total:
    print(r, total[r])

# DICT COMPREHENSION WITH .COUNT, OPTIMIZED

names = ["Jessica", "John", "Steve", "John", "John", "Steve"]
total = {name: names.count(name) for name in set(names)} # using set avoids recomputing the expensive .count more than once per unique name
    
sorted_total = sorted(total, key=total.get, reverse=True)
    
for r in sorted_total:
    print(r, total[r])

# SIMPLE LOOP WITHOUT .COUNT

names = ["Jessica", "John", "Steve", "John", "John", "Steve"]
total = {}
    
for name in names:
    if name not in total:
        total[name] = 0
    total[name]  = 1
    
sorted_total = sorted(total, key=total.get, reverse=True)
    
for r in sorted_total:
    print(r, total[r])

# LOOP WITH .GET()

names = ["Jessica", "John", "Steve", "John", "John", "Steve"]
total = {}
    
for name in names:
    total[name] = total.get(name, 0)   1
    
sorted_total = sorted(total, key=total.get, reverse=True)
    
for r in sorted_total:
    print(r, total[r])

# LOOP WITH .SETDEFAULT

names = ["Jessica", "John", "Steve", "John", "John", "Steve"]
total = {}
    
for name in names:
    total.setdefault(name,0)
    total[name]  = 1
    
sorted_total = sorted(total, key=total.get, reverse=True)
    
for r in sorted_total:
    print(r, total[r])

# LOOP WITH DEFAULTDICT

import collections # defaultdict

names = ["Jessica", "John", "Steve", "John", "John", "Steve"]
total = collections.defaultdict(int)
    
for name in names:
    total[name]  = 1
    
sorted_total = sorted(total, key=total.get, reverse=True)
    
for r in sorted_total:
    print(r, total[r])

# COUNTER

import collections # Counter

names = ["Jessica", "John", "Steve", "John", "John", "Steve"]
total = collections.Counter(names)
    
sorted_total = total.most_common()
    
for name,nb in sorted_total:
    print(name, nb)

uj5u.com熱心網友回復：

您可能需要：

sorted_total = dict(sorted(zip(total, map(total.get, total)), key=lambda x: x[1], reverse=True))

標籤：Python 排序

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