Question

我想减少此词典列表以获取重复项的最新记录，其中重复项由相同的 project_name 和相同的 feature_group_name 确定。我该怎么做呢？

我现在的操作方式如下，但是我敢肯定，没有熊猫的需求还有更好的方法：

from stack import stack        
def getAction():
        '''
        Write docstring to describe function
        Inputs: no arguments taken by this function.
        Returns: string containing users input if he has chosen one of the correct input options
        '''
        correct_input=False
        while correct_input==False:
            user_input=input("Enter = to enter a URL, < to go back, > to go forward, q to quit: ")
            if user_input =='q' or user_input =='>' or user_input =='<' or user_input =='=':
                correct_input=True

            if correct_input==False:
                print('Invalid entry.')
        return user_input


    def goToNewSite(current, pages):
        '''
        Write docstring to describe function
        Inputs: index of the current website (int), reference to list containing the webpage addresses to go back and forth between
        Returns: address inputted by user as a string
        '''

        new_web_address=input('Enter a new website address ')
        for i in range(current+1,len(pages)):
            pages.pop()
        pages.append(new_web_address)
        #pages[current+1]=new_web_address
        #pages=pages[:current+1]

        return current+1



    def goBack(current, pages):
        '''
        Write docstring to describe function
        Inputs: index of the current website (int),reference to list containing the webpage addresses to go back and forth between
        Returns: index of the previous webpage (int)
        '''
        # alternatively this could be done by checking if current-1>=0
        if current-1>=0:
            return current-1

        else:
            print('Cannot go backward')
            return current




    def goForward(current, pages):
        '''
        Write docstring to describe function
        Inputs: index of the current website (int),reference to list containing the webpage addresses to go back and forth between
        Returns: index of the previous webpage (int)
        '''  
        # alternatively this could be done by checking if current +1 in range(len(pages))
        if current+1<len(pages):
            return current+1

        else:
            print('Cannot go forward')
            return current




    def main():
        '''
        Controls main flow of web browser simulator
        Inputs: N/A
        Returns: None
        '''    
        HOME = 'www.google.ca'
        websites = [HOME]
        currentIndex = 0
        quit = False

        while not quit:
            print('\nCurrently viewing', websites[currentIndex])
            print(websites)

            action = getAction()

            if action == '=':
                currentIndex = goToNewSite(currentIndex, websites)

            elif action == '<':
                currentIndex = goBack(currentIndex, websites)
            elif action == '>':
                currentIndex = goForward(currentIndex, websites)
            elif action == 'q':
                quit = True

        print('Browser closing...goodbye.')    


    if __name__ == "__main__":
        main()

示例：

class Stack:
    def __init__(self):
        self.items = []

    def push(self, item):
        self.items.append(item)

    # MODIFY: RAISE AN EXCEPTION IF THIS METHOD IS INVOKED ON AN EMPTY STACK
    def pop(self):
        if self.isEmpty()==True:
            raise Exception('Stack is empty cannot pop')
        if self.isEmpty()==False:

            return self.items.pop()

    # MODIFY: RAISE AN EXCEPTION IF THIS METHOD IS INVOKED ON AN EMPTY STACK
    def peek(self):
        if self.isEmpty()==True:
            raise Exception('Stack is empty cannot peek')
        if self.isEmpty()==False:


            return self.items[len(self.items)-1] 

    def isEmpty(self):
        return self.items == []

    def size(self):
        return len(self.items)

    def show(self):
        print(self.items)

    def __str__(self):
        stackAsString = ''
        for item in self.items:
            stackAsString += item + ' '
        return stackAsString

所需结果：

d = pd.DataFrame(l)
sorted_df = d.sort_values('datetime', ascending=False).drop_duplicates(['project_name', 'feature_group_name'],
                                                                       keep='first')
sorted_df.to_dict(orient='records')

Answer 1

    l  = [i for n, i in enumerate(l) if i not in l[n + 1:]]

Python：基于DateTime键从字典列表中删除重复项

1 个答案: