I'm learning about deep learning and start to follow a tutorial.
At the frist part when I try to run:
train_lmdb = '~/deeplearning-cats-dogs-tutorial/input/train_lmdb'
validation_lmdb = '~/deeplearning-cats-dogs-tutorial/input/validation_lmdb'
train_data = [img for img in glob.glob("../input/train/*jpg")]
test_data = [img for img in glob.glob("../input/test1/*jpg")]
#Shuffle train_data
random.shuffle(train_data)
print 'Creating train_lmdb'
in_db = lmdb.open(train_lmdb, map_size=int(1e12))
I got the erro:
OSError: [Errno 2] No such file or directory: '~/deeplearning-cats-dogs-tutorial/input/train_lmdb'
lmdb.open was supposed to create the file, right? I dont understand why it its trying to open it.
I just replace "~/" with "/home/user/" and it is working fine
Related
I'm working on object detection with YOLOV5. the last, when I saved the images and labels in the text in the folder, I got an Error. FileNotFoundError: [Errno 2] No such file or directory: 'data_images\\1 (1).jpg'
def save_data(filename, folder_path, group_obj):
src = os.path.join('data_images',filename)
dst = os.path.join(folder_path,filename)
move(src,dst) # move image to the destination folder
# save the labels
text_filename = os.path.join(folder_path, os.path.splitext(filename)[0]+'.txt')
group_obj.get_group(filename).set_index('filename').to_csv(text_filename,sep=' ',index=False,header=False)
How to solve. I checked the data is present.
I am wanting to create a file and save it to json format. Every example I find specifies the 'open' method. I am using Python 2.7 on Windows. Please help me understand why the 'open' is necessary for a file I am saving for the first time.
I have read every tutorial I could find and researched this issue but with no luck still. I do not want to create the file outside of my program and then have my program overwrite it.
Here is my code:
def savefile():
filename = filedialog.asksaveasfilename(initialdir =
"./Documents/WorkingDirectory/",title = "Save file",filetypes = (("JSON
files","*.json"), ("All files", "*.")))
with open(filename, 'r+') as currentfile:
data = currentfile.read()
print (data)
Here is this error I get:
Exception in Tkinter callback Traceback (most recent call last):
File "C:\Python27\lib\lib-tk\Tkinter.py", line 1542, in call
return self.func(*args) File "C:\Users\CurrentUser\Desktop\newproject.py", line 174, in savefile
with open(filename, 'r+') as currentfile: IOError: [Errno 2] No such file or directory:
u'C:/Users/CurrentUser/Documents/WorkingDirectory/test.json'
Ok, I figured it out! The problem was the mode "r+". Since I am creating the file, there is no need for read and write, just write. So I changed the mode to 'w' and that fixed it. I also added the '.json' so it would be automatically added after the filename.
def savefile():
filename = filedialog.asksaveasfilename(initialdir =
"./Documents/WorkingDirectory/",title = "Save file",filetypes = (("JSON
files","*.json"), ("All files", "*.")))
with open(filename + ".json", 'w') as currentfile:
line1 = currentfile.write(stringone)
line2 = currentfile.write(stringtwo)
print (line1,line2)
I created a ms-word document using MailMerge in django. It´s worked ok.
Right now, i´d like to show this file on screen. I write the code bellow, but it didn´t work.
views.py
with open(file_path) as doc:
response = HttpResponse(doc.read(), content_type='application/ms-word')
response = HttpResponse(template_output)
response['Content-Disposition'] = 'attachment;filename=var_nomdocumento_output'
error:
[Errno 13] Permission denied: 'C:\\GROWTHTECH\\Projetos\\blockchain\\media_root/procuracao'
You forgot to provide the binary open mode. It can be r open for reading (default) w open for writing, truncating the file first, b for binary mode.
so In our case: It will be rb
file_path = 'path/path/file.docx'
with open(file_path,'rb') as doc:
response = HttpResponse(doc.read(), content_type='application/ms-word')
# response = HttpResponse(template_output)
response['Content-Disposition'] = 'attachment;filename=name.docx'
return response
No browsers currently render Word Documents as far as I know. So your file will be automatically downloaded whatever the parameter is: 'attachment;filename="file_name"' or 'inline;filename="file_name"'
Using pete warden tutorials i had trained the inception network and training of which i am getting two files
1.retrained_graph.pb
2.retrained_label.txt
Using this i wanted to classify the flower image.
I had install pycharm and linked all the tensorflow library , i had also test the sample tensorflow code it is working fine.
Now when i run the label_image.py program which is
import tensorflow as tf, sys
image_path = sys.argv[1]
# Read in the image_data
image_data = tf.gfile.FastGFile(image_path, 'rb').read()
# Loads label file, strips off carriage return
label_lines = [line.rstrip() for line
in tf.gfile.GFile("/tf_files/retrained_labels.txt")]
# Unpersists graph from file
with tf.gfile.FastGFile("/tf_files/retrained_graph.pb", 'rb') as f:
graph_def = tf.GraphDef()
graph_def.ParseFromString(f.read())
_ = tf.import_graph_def(graph_def, name='')
with tf.Session() as sess:
# Feed the image_data as input to the graph and get first prediction
softmax_tensor = sess.graph.get_tensor_by_name('final_result:0')
predictions = sess.run(softmax_tensor, \
{'DecodeJpeg/contents:0': image_data})
# Sort to show labels of first prediction in order of confidence
top_k = predictions[0].argsort()[-len(predictions[0]):][::-1]
for node_id in top_k:
human_string = label_lines[node_id]
score = predictions[0][node_id]
print('%s (score = %.5f)' % (human_string, score))
i am getting this error message
/home/chandan/Tensorflow/bin/python /home/chandan/PycharmProjects/tf/tf_folder/tf_files/label_image.py
Traceback (most recent call last):
File "/home/chandan/PycharmProjects/tf/tf_folder/tf_files/label_image.py", line 7, in <module>
image_path = sys.argv[1]
IndexError: list index out of range
Could any one please help me with this issue.
You are getting this error because it is expecting image name (with path) as an argument.
In pycharm go to View->Tool windows->Terminal.
It is same as opening separate terminal. And run
python label_image.py /image_path/image_name.jpg
You are trying to get the command line argument by calling sys.argv[1]. So you need to give command line arguments to satisfy it. Looks like the argument required is a test image, you should pass its location as a parameter.
Pycharm should have a script parameters and interpreter options dialog which you can use to enter the required parameters.
Or you can call the script from a command line and enter the parameter via;
>python my_python_script.py my_python_parameter.jpg
EDIT:
According to the documents (I don't have pycharm installed on this computer), you should go to Run/Debug configuration menu and edit the configurations for your script. Add the absolute path of your file into Script Parameters box in quotes.
Or alternatively if you just want to skip the parameter thing completely, just get the path as raw_input (input in python3) or just simply give it to image_path = r"absolute_image_path.jpg"
I am new to Python and seem to be having trouble getting an image to download and save to a file. I was wondering if someone could point out my error. I have tried two methods in various ways to no avail. Here is my code below:
# Ask user to enter URL
url= http://hosted.ap.org/dynamic/stories/A/AF_PISTORIUS_TRIAL?SITE=AP&SECTION=HOME&TEMPLATE=DEFAULT&CTIME=2014-04-15-15-48-52
timestamp = datetime.date.today()
soup = BeautifulSoup(urllib2.urlopen(url).read())
#soup = BeautifulSoup(requests.get(url).text)
# ap
links = soup.find("td", {'class': 'ap-mediabox-td'}).find_all('img', src=True)
for link in links:
imgfile = open('%s.jpg' % timestamp, "wb")
link = link["src"].split("src=")[-1]
imgurl = "http://hosted.ap.org/dynamic/files" + link
download_img = urllib2.urlopen(imgurl).read()
#download_img = requests.get(imgurl, stream=True)
#imgfile.write(download_img.content)
imgfile.write(download_img)
imgfile.close()
# link outputs: /photos/F/f5cc6144-d991-4e28-b5e6-acc0badcea56-small.jpg
# imgurl outputs: http://hosted.ap.org/dynamic/files/photos/F/f5cc6144-d991-4e28-b5e6-acc0badcea56-small.jpg
I receive no console error, just an empty picture file.
The relative path of the image can be obtained as simply as by doing:
link = link["src"]
Your statement:
link = link["src"].split("src=")[-1]
is excessive. Replace it with above and you should get the image file created. When I tried it out, I could get the image file to be created. However, I was not able to view the image. It said, the image was corrupted.
I have had success in the past doing the same task using python's requests library using this code snippet:
r = requests.get(url, stream=True)
if r.status_code == 200:
with open('photo.jpg', 'wb') as f:
for chunk in r.iter_content():
f.write(chunk)
f.close()
url in the snippet above would be your imgurl computed with the changes I suggested at the beginning.
Hope this helps.