dic = {"abc":18,"adc":19,"abe":20} # 默認(rèn)對鍵排序，從小到大，返回排序后鍵組成的列表 zidian = sorted(dic)#['abc', 'abe', 'adc'] print(zidian) # 對鍵進(jìn)行反向（從大到小）排序 zidian = sorted(dic,reverse=True)#['adc', 'abe', 'abc'] print(zidian) # 拿到所有的key，然后再對key排序 zidian = sorted(dic.keys(),reverse=True)#['adc', 'abe', 'abc'] print(zidian) # 對值排序，從小到大 print(dic) zidian = sorted(dic.values())#[18, 19, 20] print(zidian) # 對值排序，從大到小 zidian = sorted(dic.values(),reverse=True)#[20, 19, 18] print(zidian) #可以用dict1.items()，得到包含鍵，值的元組, # 由于迭代對象是元組，返回值自然是元組組成的列表,x指元組，x[1]是值，x[0]是鍵# 鍵由小到大排序 zidian = sorted(dic.items(),key=lambda x:x[0]) print(zidian)# 鍵由大到小排序 zidian = sorted(dic.items(),key=lambda x:x[0],reverse=True) print(zidian)# 值由小到大排序 zidian = sorted(dic.items(),key=lambda x:x[1]) print(zidian) # 值由大到小排序 zidian = sorted(dic.items(),key=lambda x:x[1],reverse=True) print(zidian)#itemgetter(0)，獲取key # itemgetter(1)，獲取value from operator import itemgetter d = {"a":8,"b":4,"c":12,"a":10,"b":1,"e":10} # 鍵由大到小 print(sorted(d.items(),key=itemgetter(0),reverse=True))#[('e', 10), ('c', 12), ('b', 1), ('a', 10)] # 值由大到小 print(sorted(d.items(),key=itemgetter(1),reverse=True))#[('c', 12), ('a', 10), ('e', 10), ('b', 1)]

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prices = {'ACME': 45.23,'AAPL': 612.78,'IBM': 205.55,'HPQ': 37.20,'FB': 10.75 } # 在一個字典上執(zhí)行普通的數(shù)學(xué)運算，你會發(fā)現(xiàn)它們僅僅作用于鍵，而不是值 min(prices) # Returns 'AAPL' max(prices) # Returns 'IBM' min(prices.values()) # Returns 10.75 max(prices.values()) # Returns 612.78 min(prices, key=lambda k: prices[k]) # Returns 'FB' max(prices, key=lambda k: prices[k]) # Returns 'AAPL' min_price = min(zip(prices.values(), prices.keys())) # min_price is (10.75, 'FB') print("min_price",min_price) max_price = max(zip(prices.values(), prices.keys())) # max_price is (612.78, 'AAPL') print("max_price",max_price)#值由大到小排序 prices_sorted = sorted(zip(prices.values(), prices.keys()),reverse=True) # prices_sorted is [(10.75, 'FB'), (37.2, 'HPQ'), # (45.23, 'ACME'), (205.55, 'IBM'), # (612.78, 'AAPL')] print(prices_sorted)

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轉(zhuǎn)載于:https://www.cnblogs.com/songdanlee/p/11135282.html

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