python - 从列表中为 Python 字典分配和显示值的正确方法是什么？

Question

我正在读取一台设备的测量值（结果）列表，一次 20 次测量。我需要将它们分成 2 个单独的组并将它们放在 2 个不同的字典中，进行一些通过/失败计算，然后在 GUI 中显示它们。我到目前为止是这样的：

import sys

from collections import OrderedDict

from tkinter import *

x=[]
x1=[]

Results = ([1.50229780e+00, 4.94528400e-01,-4.74700000e-04,-3.42900000e- 
04,-2.24200000e-04,-1.71400000e-04,-1.71400000e-04,-1.84600000e-04,
-1.31900000e-04,-1.45100000e-04,-6.59000000e-05, 1.32000000e-05,
0.00000000e+00,-1.32000000e-05,-1.32000000e-05,-3.96000000e-05,
-1.32000000e-05,-6.59000000e-05,-3.96000000e-05,-6.59000000e-05])

# DUT1, DUT2 are dictionary....  Access by 'CH1' or 'CH9'....


DUT1 = OrderedDict()
DUT1['CH1']=0
DUT1['CH2']=0
DUT1['CH3']=0
DUT1['CH4']=0
DUT1['CH5']=0
DUT1['CH6']=0
DUT1['CH7']=0
DUT1['CH8']=0
DUT1['CH9']=0
DUT1['CH10']=0

DUT2 = OrderedDict()
DUT2['CH1']=0
DUT2['CH2']=0
DUT2['CH3']=0
DUT2['CH4']=0
DUT2['CH5']=0
DUT2['CH6']=0
DUT2['CH7']=0
DUT2['CH8']=0
DUT2['CH9']=0
DUT2['CH10']=0


i=0

for key in DUT1:
    DUT1[key]=Results[i]
    i=i+1

i=len(DUT1)

for key in DUT2:
    DUT2[key]=Results[i]
    i=i+1

root = Tk()

for key,items in DUT1.items():
    y=("%s = %s\n" % (key,items))
    sys.stdout.write("%s %s\n" % (key,items));sys.stdout.flush()
    x.append(y)

for key,items in DUT2.items():
    y1=("%s = %s\n" % (key,items))
    sys.stdout.write("%s %s\n" % (key,items));sys.stdout.flush()
    x1.append(y1)

Label(root, text = "DUT1 Results").grid(row=0,ipadx = 10,ipady = 10)
Label(root, text = x ).grid(row=1,ipadx = 10,ipady =10)

Label(root, text = "DUT2 Results").grid(row=2,ipadx = 10,ipady = 10)
Label(root, text = x1 ).grid(row=3,ipadx = 10,ipady =10)

mainloop()

我得到的输出（打印到控制台）是：

CH1 1.5022978
CH2 0.4945284
CH3 -0.0004747
CH4 -0.0003429
CH5 -0.0002242
CH6 -0.0001714
CH7 -0.0001714
CH8 -0.0001846
CH9 -0.0001319
CH10 -0.0001451

CH1 -6.59e-05
CH2 1.32e-05
CH3 0.0
CH4 -1.32e-05
CH5 -1.32e-05
CH6 -3.96e-05
CH7 -1.32e-05
CH8 -6.59e-05
CH9 -3.96e-05
CH10 -6.59e-05

为什么 DUT1 是非科学记数法，而 DUT2 是？

当标签更新时，我得到：

{CH1 = 1.5022978
}{CH2 = 0.4945284
}{CH3 = -0.0004747
}{CH4 = -0.0003429
}{CH5 = -0.0002242
}{CH6 = -0.0001714
}{CH7 = -0.0001714
}{CH8 = -0.0001846
}{CH9 = -0.0001319
}{CH10 = -0.0001451
}

{CH1 = -6.59e-05
}{CH2 = 1.32e-05
}{CH3 = 0.0
}{CH4 = -1.32e-05
}{CH5 = -1.32e-05
}{CH6 = -3.96e-05
}{CH7 = -1.32e-05
}{CH8 = -6.59e-05
}{CH9 = -3.96e-05
}{CH10 = -6.59e-05
}

为什么要打印括号？

标签： pythondictionaryuser-interface

Answer 1

Python 3 以科学计数法数字 < 1e-4 打印。例子：

print(1e-4) #out: 0.0001
print(1e-5) #out: 1e-05

您可以使用格式化打印强制执行格式

# for non-scientific:
print("{0:0.6f}".format(1e-5)) #out: 0.000010
# forcing scientific notation:
print('%.2E' % 40800000000.00000000000000) #out: 4.08E+10
print('%.4E' % 1e-2) #out: 1.0000E-02

取自这个答案的科学记数法示例。

这是您的代码的工作版本，我认为它可以满足您的需求。它通过 for 循环中的连接创建一个格式化的字符串，然后再创建一个标签来放置它。

import sys
from collections import OrderedDict
from tkinter import *

x=[]
x1=[]

Results = ([1.50229780e+00, 4.94528400e-01,-4.74700000e-04,-3.42900000e-04,-2.24200000e-04,-1.71400000e-04,-1.71400000e-04,-1.84600000e-04,-1.31900000e-04,-1.45100000e-04,-6.59000000e-05, 1.32000000e-05,0.00000000e+00,-1.32000000e-05,-1.32000000e-05,-3.96000000e-05,-1.32000000e-05,-6.59000000e-05,-3.96000000e-05,-6.59000000e-05])

# DUT1, DUT2 are dictionary....  Access by 'CH1' or 'CH9'....

DUT1 = OrderedDict()
DUT1['CH1']=0
DUT1['CH2']=0
DUT1['CH3']=0
DUT1['CH4']=0
DUT1['CH5']=0
DUT1['CH6']=0
DUT1['CH7']=0
DUT1['CH8']=0
DUT1['CH9']=0
DUT1['CH10']=0

DUT2 = OrderedDict()
DUT2['CH1']=0
DUT2['CH2']=0
DUT2['CH3']=0
DUT2['CH4']=0
DUT2['CH5']=0
DUT2['CH6']=0
DUT2['CH7']=0
DUT2['CH8']=0
DUT2['CH9']=0
DUT2['CH10']=0


i=0

for key in DUT1:
    DUT1[key]=Results[i]
    i=i+1

i=len(DUT1)

for key in DUT2:
    DUT2[key]=Results[i]
    i=i+1

root = Tk()

for key,items in DUT1.items():
    y=("%s = %s\n" % (key,items))
    sys.stdout.write("%s %.4E\n" % (key,items));sys.stdout.flush()
    x.append(y)

for key,items in DUT2.items():
    y1=("%s = %s\n" % (key,items))
    sys.stdout.write('%s %.4E\n' % (key,items));sys.stdout.flush()
    x1.append(y1)

Label(root, text = "DUT1 Results").grid(row=0,ipadx = 10,ipady = 10)
s = ''
for key,items in DUT1.items():
    s += '%s %.4E\n' % (key,items)
Label(root, text = s ).grid(row=1,ipadx = 10,ipady =10)

Label(root, text = "DUT2 Results").grid(row=2,ipadx = 10,ipady = 10)
s1 = ''
for key,items in DUT2.items():
    s1 += '%s %.4E\n' % (key,items)
Label(root, text = s1 ).grid(row=3,ipadx = 10,ipady =10)

mainloop()