Question

我有一个关于改组的问题，但首先，这是我的代码：

from psychopy import visual, event, gui
import random, os
from random import shuffle
from PIL import Image
import glob
a = glob.glob("DDtest/targetimagelist1/*")
b = glob.glob("DDtest/distractorimagelist1/*")
target = a
distractor = b
pos1 = [-.05,-.05]
pos2 = [.05, .05]

shuffle(a)
shuffle(b)

def loop_function_bro():
    win = visual.Window(size=(1280, 800), fullscr=True, screen=0, monitor='testMonitor', color=[-1,-1,-1], colorSpace='rgb')
        distractorstim = visual.ImageStim(win=win,
        image= distractor[i], mask=None,
        ori=0, pos=pos1, size=[0.5,0.5],
        color=[1,1,1], colorSpace='rgb', opacity=1,
        flipHoriz=False, flipVert=False,
        texRes=128, interpolate=True, depth=-1.0)

    targetstim= visual.ImageStim(win=win,
        image= target[i], mask=None,
        ori=0, pos=pos2, size=[0.5,0.5],
        color=[1,1,1], colorSpace='rgb', opacity=1,
        flipHoriz=False, flipVert=False,
        texRes=128, interpolate=True, depth=-2.0)

    distractorstim.setAutoDraw(True)
    targetstim.setAutoDraw(True)
    win.flip()
    event.waitKeys(keyList = ['space'])

for i in range (2):  
    loop_function_bro()

此代码随机地移动一堆图像并显示它们。但是，我希望它以相同的顺序对图像进行洗牌，以便两个列表以相同的随机顺序显示。有办法做到这一点吗？

干杯，:)

Answer 1

我能想到的最简单的方法是使用单独的索引列表，而shuffle代替它。

indices = list(xrange(len(a)))  # Or just range(len(a)) in Python 2
shuffle(indices)

Answer 2

此问题已获得回复here和here。我特别喜欢以下语法的简单性

randomizedItems = map(items.__getitem__, indices)

有关完整的工作示例，请参阅下面的代码。请注意，我已经改变了很多，使代码更短，更清晰，更快。见评论。

# Import stuff. This section was simplified.
from psychopy import visual, event, gui  # gui is not used in this script
import random, os, glob
from PIL import Image

pos1 = [-.05,-.05]
pos2 = [.05, .05]

# import two images sequences and randomize in the same order
target = glob.glob("DDtest/targetimagelist1/*")
distractor = glob.glob("DDtest/distractorimagelist1/*")
indices = random.sample(range(len(target)), len(target))  # because you're using python 2
target = map(target.__getitem__, indices)
distractor = map(distractor.__getitem__, indices)

# Create window and stimuli ONCE. Think of them as containers in which you can update the content on the fly.
win = visual.Window(size=(1280, 800), fullscr=True, screen=0, monitor='testMonitor', color=[-1,-1,-1])  # removed a default value
targetstim = visual.ImageStim(win=win, pos=pos2, size=[0.5,0.5])
targetstim.autoDraw = True
distractorstim = visual.ImageStim(win=win, pos=pos1, size=[0.5,0.5])  # removed all default values and initiated without an image. Set autodraw here since you didn't seem to change it during runtime. But feel free to do it where you please.
distractorstim.autoDraw = True

# No need to pack in a function. Just loop immediately. Rather than just showing two stimuli, I've set it to loop over all stimuli.
for i in range(len(target)):
    # set images. OBS: this may take some time. Probably between 20 and 500 ms depending mainly on image dimensions. Smaller is faster. It's still much faster than generating a full Window/ImageStim each loop though.
    distractorstim.image = distractor[i]
    targetstim.image = target[i]

    # Display and wait for answer
    win.flip()
    event.waitKeys(keyList = ['space'])

在保持配对的同时随机化图像的另一种方法是你可能在某些时候做的事情：把它们放在一个字典列表中，每个字典代表一个试验。因此，而不是两条map行，请执行：

trialList = [{'target':target[i], 'distractor':distractor[i]} for i in indices]

尝试打印试用列表（print trialList）以查看其外观。然后循环试验：

for trial in trialList:
    # set images.
    targetstim.image = trial['target']
    distractorstim.image = trial['distractor']

    # Display and wait for answer. Let's record reaction times in the trial, just for fun.
    win.flip()
    response = event.waitKeys(keyList = ['space'], timeStamped=True)
    trial['rt'] = response[0][1]  # first answer, second element is rt.

Answer 3

我会创建一个数字数组，对其进行随机排序，并根据随机数字对图像列表进行排序。

因此两个列表都以同样的方式进行洗牌。

Answer 4

    data1=[foo bar foo]
    data2=[bar foo bar]
    data3=[foo bar foo]
    alldata=zip((data1,data2,data3))
# now do your shuffling on the "outside" index of alldata then
    (data1shuff,data2shuff,data3shuff) = zip(*alldata)

在python中以相同的顺序随机播放两个列表

4 个答案: