有這樣的串列： ["Apple", "Banana", "Peach", "Apple", "Banana", "Orange"]
我試圖將串列傳遞給一個函式并按如下方式對其進行排序：
def l(fruits):
for i in fruits:
print(i)
n = ["Apple", "Banana", "Peach", "Apple", "Banana", "Orange"]
l(sorted(n))
我只得到這樣的結果：
Apple
Apple
Banana
Banana
Orange
Peach
我想知道如何分別計算串列中每個專案的出現次數以獲得如下輸出：
Apple 2
Banana 2
Orange 1
Peach 1
我知道使用collections.Counter可以獲得類似的結果，但是，我希望輸出與上面相同。
編輯：我剛剛發現串列中的專案應該是小寫的。
另外，我嘗試在沒有 collections_counter
def fruit_list(fruits):
for i in fruits:
fruit_occurence = fruits.count("apple")
fruit_occurence2 = fruits.count("banana")
fruit_occurence3 = fruits.count("peach")
fruit_occurence4 = fruits.count("orange")
if i == "apple":
print(i.title(), *{fruit_occurence})
elif i == "banana":
print(i.title(), *{fruit_occurence2})
elif i == "peach":
print(i.title(), *{fruit_occurence3})
elif i == "orange":
print(i.title(), *{fruit_occurence4})
f = ["apple", "banana", "peach", "apple", "banana", "orange"]
fruit_list(sorted(f))
輸出：
Apple 2
Apple 2
Banana 2
Banana 2
Orange 1
Peach 1
無論如何要洗掉輸出中的重復項？將串列更改為 set 會影響fruit_occurances。
uj5u.com熱心網友回復：
您可以使用Counter，然后使用sorted在for中function，然后用print(*...)象下面這樣：
from collections import Counter
def l(fruits):
for i in sorted(Counter(fruits).items()):
print(*i)
n = ["Apple", "Banana", "Peach", "Apple", "Banana", "Orange"]
l(n)
輸出：
Apple 2
Banana 2
Orange 1
Peach 1
uj5u.com熱心網友回復：
使用collections.Counter：
from collections import Counter
def l(fruits):
c = Counter(fruits)
for fruit, count in sorted(c.items()):
print(fruit, count)
uj5u.com熱心網友回復：
完全不用修改l，
>>> l(f'{f} {c}' for f, c in sorted(Counter(n).items()))
訣竅是創建一個生成器運算式，該運算式生成要從Counter實體列印的字串。
uj5u.com熱心網友回復：
text = ["Apple", "Banana", "Peach", "Apple", "Banana", "Orange"]
dict = {}
def l(text):
for i in text:
dict[i] = text.count(i)
print(dict)
l(text)
uj5u.com熱心網友回復：
這應該適合你
import collections
fruits = ["Apple", "Banana", "Peach", "Apple", "Banana", "Orange"]
d = {}
for fruit in fruits:
if fruit in d:
d[fruit] = 1
else:
d[fruit] = 1
od = collections.OrderedDict(sorted(d.items()))
print (od)
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