Question

我有一个字典列表，如下所示：

data = [{'Name': 'Paul', 'Date': '20200412', 'ID': '1020'}, {'Name': 'Frank', 'Date': '20200413', 'ID': '1030'}, {'Name': 'Anna', 'Date': '20200414', 'ID': '1040'}]

我需要创建一个新的词典列表，其中ID的值为键，而值是另一个具有与此特定ID关联的键/值的字典。这是所需的输出：

new_data = [{'1020': {'Name': 'Paul', 'Date': '20200412'}},
{'1030': {'Name': 'Frank', 'Date': '20200413'}},
{'1040': {'Name': 'Anna', 'Date': '20200414'}}]

我尝试过：

for index, my_dict in enumerate(data):
    new_data = []
    key = my_dict['ID']
    new_data.append(key)

但是那只分配了键值，不确定如何将其与其他键/值一起推入新的字典中。

Answer 1

>>> [{i['ID']: {k:v for k,v in i.items() if k != 'ID'}} for i in data]
[{'1020': {'Name': 'Paul', 'Date': '20200412'}},
 {'1030': {'Name': 'Frank', 'Date': '20200413'}},
 {'1040': {'Name': 'Anna', 'Date': '20200414'}}]

Answer 2

您可以尝试以下列表理解：

X = "X"
O = "O"
EMPTY = None

def initial_state():
    """
    Returns starting state of the board.
    """
    return [[EMPTY, EMPTY, EMPTY],
            [EMPTY, EMPTY, EMPTY],
            [EMPTY, EMPTY, EMPTY]]


def player(board):
    """
    Returns player who has the next turn on a board.
    """
    o_counter = 0
    x_counter = 0
    for i in board:
        for j in i:
            if j == 'X':
                x_counter += 1
            elif j == 'O':
                o_counter += 1
    if x_counter == 0 and o_counter == 0:
        return 'O'
    elif x_counter > o_counter:
        return 'O'
    elif o_counter > x_counter:
        return 'X'



def actions(board):
    """
    Returns set of all possible actions (i, j) available on the board.
    """
    action = []
    for i in range(3):
        for j in range(3):
            if board[i][j] is None:
                action.append([i, j])
    return action


def result(board, action):
    """
    Returns the board that results from making move (i, j) on the board.
    """
    p = player(board)
    i, j = action
    board[i][j] = p
    return board


def winner(board):
    """
    Returns the winner of the game, if there is one.
    """
    i = 1
    if board[0][0] == board[1][1] == board[2][2] and (board[0][0] == 'X' or board[0][0] == 'O'):
        return board[0][0]
    elif board[0][2] == board[1][1] == board[2][0] and (board[0][2] == 'X' or board[0][2] == 'O'):
        return  board[0][2]
    else:
        if board[0][0] == board[0][1] == board[0][2] and (board[0][0] == 'X' or board[0][0] == 'O'):
            return board[0][0]
        elif board[i][0] == board[i][1] == board[i][2] and (board[i][0] == 'X' or board[i][0] == 'O'):
            return board[i][0]
        elif board[2][0] == board[2][1] == board[2][2] and (board[2][0] == 'X' or board[2][0] == 'O'):
            return board[2][0]

        elif board[0][0] == board[1][0] == board[2][0] and (board[0][0] == 'X' or board[0][0] == 'O'):
            return board[0][0]
        elif board[0][i] == board[1][i] == board[2][i] and (board[0][i] == 'X' or board[0][i] == 'O'):
            return board[0][i]
        elif board[0][2] == board[1][2] == board[2][2] and (board[0][2] == 'X' or board[0][2] == 'O'):
            return board[0][2]

def terminal(board):
    """
    Returns True if game is over, False otherwise.
    """
    check = True
    if winner(board) == 'X' or winner(board) == 'O':
        return True
    elif check:
        for i in board:
            for j in i:
                if j is None:
                    check = False
                    return False
        if check:
            return True
    else:
        return False


def utility(board):
    """
    Returns 1 if X has won the game, -1 if O has won, 0 otherwise.
    """
    if winner(board) == 'X':
        return 1
    elif winner(board) == 'O':
        return -1
    else:
        return 0


def maximum(board):
    if terminal(board):
        return utility(board)
    v = -9999999999999999999999
    for action in actions(board):
        m = minimum(result(board, action))
        if m > v:
            v = m
    return v


def minimum(board):
    if terminal(board):
        return utility(board)
    v = 9999999999999999999999
    for action in actions(board):
        m = maximum(result(board, action))
        if m < v:
            v = m
    return v


def minimax(board):
    """
    Returns the optimal action for the current player on the board.
    """
    return_action = None
    curr_player = player(board)
    states = actions(board)
    temp_board = board.copy()
    score = 0
    temp_score = 0
    for state in states:
        i, j = state
        if curr_player == 'X':
            temp_board[i][j] = curr_player
            temp_score = maximum(temp_board)
        elif curr_player == 'O':
            temp_board[i][j] = curr_player
            temp_score = minimum(temp_board)
        if curr_player == 'X':
            if temp_score > score:
                score = temp_score
                return_action = state
        elif curr_player == 'O':
            if temp_score < score:
                score = temp_score
                return_action = state

    return  return_action

其中将[{x["ID"]: {k: v for k, v in x.items() if k != "ID"}} for x in data]作为字典的父键，并在字典理解中从子字典中过滤掉ID键

可以细分为以下哪个：

ID

甚至是直接循环的方法：

result = []
for x in data:
    result.append({x["ID"]: {k: v for k, v in x.items() if k != "ID"}})

输出：

result = []
for x in data:
    dic = {x["ID"]: {}}
    for k, v in x.items():
        if k != "ID":
            dic[x["ID"]][k] = v
    result.append(dic)

Answer 3

new_data = []
for index, my_dict in enumerate(data):
    key = my_dict['ID']
    del my_dict['ID']
    new_data.append({key:data[index]})

Answer 4

要添加其他值，您只需要创建一个新的字典，像这样：

new_data.append( key:{
'name':my_dict['name']
'Date':my_dict['date']
}

您也不需要设置'key'变量，只需使用'my_dict ['ID']'

如何从另一个键Python的值创建新键

4 个答案: