蟒蛇列表排序

Question

my_list=raw_input('Please enter a list of items (separated by comma): ') 
my_list=my_list.split() 
my_list.sort() 

print "List statistics: " 

for x in my_list: 
    z=my_list.count(x) 

    if z>1: 
     print x, "is repeated", z, "time." 
    else: 
     print x, "is repeated", z, "times." 

我想讓程序按字母順序排序列表，然後打印它找到的每一個。輸出是：

List statistics: 
bird, is repeated 1 time. 
cat, is repeated 1 time. 
dog is repeated 1 time. 
dog, is repeated 2 times. 
dog, is repeated 2 times. 

我只需要它打印一次。此外，我想弄清楚如何把項目放在引號中。

Answer 1

可以遍歷從列表中創建一組：

for x in set(my_list): 
    z = my_list.count(x) 

這樣一來，你只會在循環得到每個元素一次。

Answer 2

更換for x in my_list:與for x in sorted(list(set(my_list))):

而且雖然在問題還沒問你可以使用這樣的：

print x, "is repeated", z, "time%s." %("s" if z > 1 else "") 

代替：

if z>1: 
    print x, "is repeated", z, "time." 
else: 
    print x, "is repeated", z, "times." 

「Python化」，因爲他們稱呼它。

Answer 3

from itertools import Counter 
my_list=raw_input('Please enter a list of items (separated by comma): ') 
my_list=my_list.split(",") 


print "List statistics: " 
import operator 
for item,count in sorted(Counter(my_list).items(),key =operator.itemgetter(0)) : 

    if z==1: 
     print '"%s" is repeated %d time.'%(item,count) 
    else: 
     print '"%s" is repeated %d times.'%(item,count) 

如果你使用python < 2.7，你可以使自己的計數器法

def Counter(a_list): 
    d = {} 
    for item in a_list: 
     if d.get(item,None) is None: 
      d[item] = 0 
     d[item] += 1 
    return d