Question

我得到了这3个字典：

self.Update()

我做了一个新的决定：

import wx


class Mywin(wx.Frame):

    def __init__(self, parent, title):
        super(Mywin, self).__init__(parent, title=title, size=(500, 300))

        self.InitUI()

    def InitUI(self):

        self.panel = wx.Panel(self)
        self.panel.SetBackgroundColour(wx.Colour('RED'))


        self.Centre()
        self.Show(True)

        menuBar = wx.MenuBar()
        RectangleButton = wx.Menu()

        Item1 = RectangleButton.Append(wx.ID_ANY, 'Rectangle 1')
        Item2 = RectangleButton.Append(wx.ID_ANY, 'Rectangle 2')

        menuBar.Append(RectangleButton, 'Rectangles')

        self.SetMenuBar(menuBar)

        self.Bind(wx.EVT_MENU, self.DrawRect1, Item1)
        self.Bind(wx.EVT_MENU, self.DrawRect2, Item2)


    def DrawRect1(self, e):
        self.panel.SetBackgroundColour(wx.Colour('BLUE'))
        self.Refresh()
        self.Update()
        self.dc = wx.ClientDC(self.panel)
        self.dc.SetBrush(wx.Brush(wx.Colour('white')))
        self.dc.DrawRectangle(10, 10, 100, 100)

    def DrawRect2(self, e):
        self.panel.SetBackgroundColour(wx.Colour('GREEN'))
        self.Refresh()
        self.Update()        
        self.dc = wx.ClientDC(self.panel)
        self.dc.SetBrush(wx.Brush(wx.Colour('white')))
        self.dc.DrawRectangle(20, 20, 50, 50)

myApp = wx.App()
Mywin(None,'Drawing demo')
myApp.MainLoop()

然后我用：

d = {'x': 1, 'y': 2, 'z': 3}

e = {'x': 2, 'y': 3, 'z': 1}

f = {'t': 1, 'q': 2, 'n': 3}

我得到了：

new_dict = {}

但是我想得到：

new_dict.update(d)
new_dict.update(e)
new_dict.update(f)

换句话说，我不想每个键和值都覆盖它的前一个-我想对每个键的值求和。我该怎么办？

Answer 1

使用collections.Counter

例如：

from collections import Counter

d = {'x': 1, 'y': 2, 'z': 3}
e = {'x': 2, 'y': 3, 'z': 1}
f = {'t': 1, 'q': 2, 'n': 3}

new_dict = Counter(d) + Counter(e) + Counter(f)
print(new_dict)

输出：

Counter({'y': 5, 'z': 4, 'n': 3, 'x': 3, 'q': 2, 't': 1})

Answer 2

尝试：

from collections import Counter

d = {'x': 1, 'y': 2, 'z': 3}
e = {'x': 2, 'y': 3, 'z': 1}
f = {'t': 1, 'q': 2, 'n': 3}

l = [d,e,f]

new_dict = Counter()
for d in l:
    new_dict.update(d)

您会得到：

Counter({'x': 3, 'y': 5, 'z': 4, 't': 1, 'q': 2, 'n': 3})

Answer 3

您可以遍历每个字典并添加相同键的值，而不用创建3个Collections.Counter实例

d = {'x': 1, 'y': 2, 'z': 3}
e = {'x': 2, 'y': 3, 'z': 1}
f = {'t': 1, 'q': 2, 'n': 3}

dct = {}
#Iterate over every dict
for l in [d, e, f]:
    #Add up values for the same key
    for k,v in l.items():
        if k in dct:
            dct[k] += v
        else:
            dct[k] = v

print(dct)
#{'x': 3, 'y': 5, 'z': 4, 't': 1, 'q': 2, 'n': 3}

字典中的键值总和在python 3中合并

3 个答案: