我想弄清楚如何跟蹤用戶輸入的列表中數字的頻率。我的意思是：假設有人輸入 45 兩次，52 次輸入 52 次，22 次輸入，代碼會打印出類似“頻率：45-2、52-3 和 22-1”的內容。提出這個問題的其他代碼是針對已經在代碼中創建的列表，但這個不同，因為用戶正在添加到列表中。import sysprint ("After inputting this data, the sum, average, maximum and minimum number will be printed")temperatureList = list()weather=int(input("Enter the amount of days that you are looking at for the weather:"))print("Enter the high temperature for those next days: ")for i in range(int(weather)): k=int(input("")) temperatureList.append(int(k))sm=sum(temperatureList)avg=sm/weatherprint("SUM = ",sm)print("AVERAGE = ",avg)temperatureList.sort()print("This is the numbers from low to high", temperatureList)print("Maximum number in the list is:", max(temperatureList), "and the minimum number is: ", min(temperatureList))while True: action = int(input("Input add if you want to add to the list again. Or remove if you want to remove from the list, or end if you want to end this program")) if action == add: print("Enter what you want to be added ") add = int(input("")) temperatureList.append(add) print(temperatureList) sm = sum(temperatureList) avg = sm / weather print("SUM = ", sm) print("AVERAGE = ", avg) temperatureList.sort() print("This is the numbers from low to high", temperatureList) elif action == remove: if len(temperatureList) > 1: del temperatureList[-1] print(temperatureList) sm = sum(temperatureList) avg = sm / weather print("SUM = ", sm) print("AVERAGE = ", avg) else: print("Everything removed from the list") elif action == end: sys.exit()
2 回答
catspeake
TA貢獻1111條經驗 獲得超0個贊
GeeksforGeeks上有一個很好的例子來說明這個問題。閱讀本文將幫助您解決問題。
這是他們為解決方案提供的 Python 代碼：
# Python program to count the frequency of
# elements in a list using a dictionary
def CountFrequency(my_list):
# Creating an empty dictionary
freq = {}
for item in my_list:
if (item in freq):
freq[item] += 1
else:
freq[item] = 1
for key, value in freq.items():
print ("% d : % d"%(key, value))
# Driver function
if __name__ == "__main__":
my_list =[1, 1, 1, 5, 5, 3, 1, 3, 3, 1, 4, 4, 4, 2, 2, 2, 2]
CountFrequency(my_list)
這只是遍歷列表，將列表的每個不同元素用作字典中的鍵，并將該鍵的相應計數存儲為值。
它的時間復雜度為 O(n)，其中 n 是列表的長度，因為它遍歷列表中的每個值。
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