使用if语句访问变量列表坐标

时间:2012-01-17 15:52:59

标签: python list if-statement tkinter

对于这个python 2.7 tkinter代码,如果我输入'Apple'并点击'Search'按钮,我应该将字符串变量选项和相关的radiobutton从未知(“?”)重置为描述苹果的那些(“Crunchy”)和(“Temperate”)但我在使用if语句访问列表坐标时遇到了问题。

from Tkinter import*

class Fruit:
    def __init__(self, parent):

    # variables
    self.texture_option = StringVar()
    self.climate_option = StringVar()

    # layout
    self.myParent = parent

    self.main_frame = Frame(parent, background="light blue")
    self.main_frame.pack(expand=YES, fill=BOTH)

    texture_options = ["Soft", "Crunchy","?"]
    climate_options = ["Temperate", "Tropical","?"]

    self.texture_option.set("?")
    self.climate_option.set("?")

    self.texture_options_frame = Frame(self.main_frame, borderwidth=3, background="light blue")
    self.texture_options_frame.pack(side=TOP, expand=YES, anchor=W)
    Label(self.texture_options_frame, text="Texture:", relief=FLAT, font="bold", background="light blue").pack(side=LEFT,anchor=W)
    for option in texture_options:
        button = Radiobutton(self.texture_options_frame, text=str(option), indicatoron=0,
        value=option, padx=5, variable=self.texture_option, background="light blue")
        button.pack(side=LEFT)

    self.climate_options_frame = Frame(self.main_frame, borderwidth=3, background="light blue")
    self.climate_options_frame.pack(side=TOP, expand=YES, anchor=W)
    Label(self.climate_options_frame, text="Climate:", relief=FLAT, font="bold", background="light blue").pack(side=LEFT,anchor=W)
    for option in climate_options:
        button = Radiobutton(self.climate_options_frame, text=str(option), indicatoron=0,
        value=option, padx=5, variable=self.climate_option, background="light blue")
        button.pack(side=LEFT)

    #search button
    self.search_frame = Frame(self.main_frame, borderwidth=5, height=50, background="light blue")
    self.search_frame.pack(expand=NO)

    self.enter = Entry(self.search_frame, width=30)
    self.enter.pack(side=LEFT, expand=NO, padx=5, pady=5, ipadx=5, ipady=5)

    self.searchbutton = Button(self.search_frame, text="Search", foreground="white", background="blue",
    width=6, padx="2m", pady="1m")
    self.searchbutton.pack(side=LEFT, pady=5)
    self.searchbutton.bind("<Button-1>", self.searchbuttonclick)
    self.searchbutton.bind("<Return>", self.searchbuttonclick)


def searchbuttonclick(self,event):
    #fruit  texture  climate 
    fruit_bowl=[
    ('Apple', 'Crunchy','Temperate'),
    ('Orange', 'Soft','Tropical'),
    ('Pawpaw','Soft','Temperate')]

    if self.enter.get()==fruit_bowl[i][0]:
        self.texture_option.set(fruit_bowl[i][1])
        self.climate_option.set(fruit_bowl[i][2])


root = Tk()
root.title("Fruit Bowl")
fruit = Fruit(root)
root.mainloop()

我想说,如果输入窗口等于fruit_bowl中任何给定行的第0列,则纹理选项设置为该行的第1列值,而climate选项设置为该行的第2列值,但是如何做我在python中说过吗?

我最初省略了这段代码的gui组件以简化操作,但显然只是使一切变得更复杂,并使我的代码看起来不稳定和奇怪。上面的代码应该为您提供一个很好的GUI窗口,但点击搜索按钮除了生成以下错误消息之外什么都不做:

Exception in Tkinter callback
Traceback (most recent call last):
File "C:\Python25\Lib\lib-tk\Tkinter.py", line 1403, in __call__
return self.func(*args)
File "F:\Python\fruit.py", line 59, in searchbuttonclick
if self.enter.get()==fruit_bowl[i][0]:
NameError: global name 'i' is not defined

是否有列表理解或我可以使用什么来解决这个而不是重写我的代码?这是一个模拟的例子,试图解决我在一个更大的模块中遇到的问题。

2 个答案:

答案 0 :(得分:4)

你应该使用字典:

fruits = {'Apple': ['Crunchy', 'Temperate'],
          'Orange': ['Soft', 'Tropical'],
          'Pawpaw': ['Soft', 'Temperate']}
print 'Apples are {}.'.format(' and '.join(fruits['Apple']))

修改:另请参阅standard library documentationofficial tutorial

编辑#2:当然你也可以像这样设置变量:

self.texture_option.set(fruits['Apple'][0])
self.climate_option.set(fruits['Apple'][1])

你也可以这样写:

fruit = self.enter.get()
self.texture_option.set(fruits.get(fruit, ['?', '?'])[0])
self.climate_option.set(fruits.get(fruit, ['?', '?'])[0])

如果您的程序不知道水果,请选择['?', '?']作为选项。

答案 1 :(得分:2)

你的示例代码不是很一致。例如,您定义了一个StingVar对象,如果您将对象与像Entry小部件这样的tkinter小部件耦合,这将是有意义的:

self.entry_var = StringVar()
self.enter = Entry(root, width = 30, textvariable = self.entry_var)
selection = self.entry_var.get()

考虑到这一点,我会遗漏你的头脑,并且这样做:

self.enter = Entry(root, width=30)
self.enter.pack(side=LEFT, expand=NO)

#fruit  texture  climate 
fruit_bowl={'apple': ('Crunchy','Temperate'),
            'orange': ('Soft','Tropical'),
            'pawpaw': ('Soft','Temperate')}

selection = self.enter.get()
try:
   self.texture_option = fruit_bowl[selection.lower()][0]
   self.climate_option = fruit_bowl[selection.lower()][1]
   self.fruit_option = selection.capitalize()
except KeyError:
    print "%s not in fruit-bowl" % selection

如果您希望保持代码不变,则必须执行以下操作:

for fruit in fruit_bowl:
    i = fruit_bowl.index(fruit)
    if self.enter.get()==fruit_bowl[i][0]:
        self.texture_option.set(fruit_bowl[i][1])
        self.climate_option.set(fruit_bowl[i][2])

您在哪里定义变量i?我看不到定义,所以也不能python。 为了纠正这种情况,我对fruit_bowl进行了迭代并分配了值 列表中的实际元组索引到变量`i。 这是您必须添加的两行(除了以下行的添加标识),以使您的代码工作。它并不优雅,但也许你可以从中学到一些东西。

另外,您也可以考虑这样做:

for i in xrange(len(fruit_bowl)):
    if self.enter.get()==fruit_bowl[i][0]:
        self.texture_option.set(fruit_bowl[i][1])
        self.climate_option.set(fruit_bowl[i][2])

如果您还有其他问题,请发表评论,我会相应更新我的答案。