I'm trying to draw a network on a basemap overlay.
I have packages:
basemap=1.3.0=py36ha7665c8_0
matplotlib=3.3.1=0
matplotlib-base=3.3.1=py36hba9282a_0
networkx=2.5=py_0
When I run only the line
from mpl_toolkits.basemap import Basemap
I get
from matplotlib.cbook import dedent
ImportError: cannot import name 'dedent'
I've tried several different versions of the packages but cannot manage to find the right functioning combination.
Anyone have any ideas on combinations of matpltlib and basemap that work? Or another way to plot my network over a basemap?
The best that worked for me is:
downgrade to :
pip install -U matplotlib==3.2
I was have the same error with importing Basemap from when I pass from python 3.2.7 to 3.3.6.
The error message comes from the fact that you try to import Basemap from mpl_toolkits.basemap, but the mpl_toolkits.basemap module requires to import the dedent function from the matplotlib.cbook module, but this function is not there.
So I guess there were two possible solution : to comment the line which import this function or to copy it. I chose the second option.
I don't know why the dedent function is not present in the matplotlib.cbook module.
Here it is the dedent funtion as I found it on the link I put, it was also have for header : #deprecated("3.1", alternative="inspect.cleandoc")
def dedent(s):
"""
Remove excess indentation from docstring *s*.
Discards any leading blank lines, then removes up to n whitespace
characters from each line, where n is the number of leading
whitespace characters in the first line. It differs from
textwrap.dedent in its deletion of leading blank lines and its use
of the first non-blank line to determine the indentation.
It is also faster in most cases.
"""
# This implementation has a somewhat obtuse use of regular
# expressions. However, this function accounted for almost 30% of
# matplotlib startup time, so it is worthy of optimization at all
# costs.
if not s: # includes case of s is None
return ''
match = _find_dedent_regex.match(s)
if match is None:
return s
# This is the number of spaces to remove from the left-hand side.
nshift = match.end(1) - match.start(1)
if nshift == 0:
return s
# Get a regex that will remove *up to* nshift spaces from the
# beginning of each line. If it isn't in the cache, generate it.
unindent = _dedent_regex.get(nshift, None)
if unindent is None:
unindent = re.compile("\n\r? {0,%d}" % nshift)
_dedent_regex[nshift] = unindent
result = unindent.sub("\n", s).strip()
return result
I copy the function dedent from the site of matplotlib : https://matplotlib.org/3.1.1/_modules/matplotlib/cbook.html#dedent inside the module init.py - matplolib\cbook.
And now it is working for me.
Be aware to copy it at the correct line, because some of its variable pre-defined in the module such as : _find_dedent_regex for the line :
match = _find_dedent_regex.match(s)
and _dedent_regex for the lines :
unindent = _dedent_regex.get(nshift, None)
if unindent is None:
unindent = re.compile("\n\r? {0,%d}" % nshift)
_dedent_regex[nshift] = unindent
This is where I put the function in the moddule
I deeply apologize for spelling and/or grammar errors that I could do, I will do my best to correct those I would have missed and reported to me.
I hope this was usefull.
Matplotlib has removed cbook.dedent since 3.3.0.
Upgrade basemap to fix this: pip install -U basemap
I tried to find a solution online that does not require altering any files. While in an Anaconda environment, use:
conda install matplotlib=3.2
Please note that support for Basemap is discontinued and should be avoided.
Related
I am trying to do a basic colour conversion in python however I can't seem to get past the below error. I have re-installed python, opencv and tried on both python 3.4.3 (latest) and python 2.7 (which is on my Mac).
I installed opencv using python's package manager opencv-python.
Here is the code that fails:
frame = cv2.imread('frames/frame%d.tiff' % count)
frame_HSV= cv2.cvtColor(frame,cv2.COLOR_RGB2HSV)
This is the error message:
cv2.error: OpenCV(3.4.3) /Users/travis/build/skvark/opencv-python/opencv/modules/imgproc/src/color.cpp:181: error: (-215:Assertion failed) !_src.empty() in function 'cvtColor'
This error happened because the image didn't load properly. So you have a problem with the previous line cv2.imread. My suggestion is :
check if the image exists in the path you give
check if the count variable has a valid number
If anyone is experiencing this same problem when reading a frame from a webcam:
Verify if your webcam is being used on another task and close it. This wil solve the problem.
I spent some time with this error when I realized my camera was online in a google hangouts group. Also, Make sure your webcam drivers are up to date
I kept getting this error too:
Traceback (most recent call last):
File "face_detector.py", line 6, in <module>
gray_img=cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
cv2.error: OpenCV(4.1.0) C:\projects\opencv-python\opencv\modules\imgproc\src\color.cpp:182: error: (-215:Assertion failed) !_src.empty() in function 'cv::cvtColor
My cv2.cvtColor(...) was working fine with \photo.jpg but not with \news.jpg. For me, I finally realized that when working on Windows with python, those escape characters will get you every time!! So my "bad" photo was being escaped because of the file name beginning with "n". Python took the \n as an escape character and OpenCV couldn't find the file!
Solution:
Preface file names in Windows python with r"...\...\" as in
cv2.imread(r".\images\news.jpg")
If the path is correct and the name of the image is OK, but you are still getting the error
use:
from skimage import io
img = io.imread(file_path)
instead of:
cv2.imread(file_path)
The function imread loads an image from the specified file and returns
it. If the image cannot be read (because of missing file, improper permissions, unsupported or invalid format), the function returns an empty matrix ( Mat::data==NULL ).
check if the image exists in the path and verify the image extension (.jpg or .png)
Check whether its the jpg, png, bmp file that you are providing and write the extension accordingly.
Another thing which might be causing this is a 'weird' symbol in your file and directory names. All umlaut (äöå) and other (éóâ etc) characters should be removed from the file and folder names. I've had this same issue sometimes because of these characters.
Most probably there is an error in loading the image, try checking directory again.
Print the image to confirm if it actually loaded or not
In my case, the image was incorrectly named. Check if the image exists and try
import numpy as np
import cv2
img = cv2.imread('image.png', 0)
cv2.imshow('image', img)
I've been in same situation as well, and My case was because of the Korean letter in the path...
After I remove Korean letters from the folder name, it works.
OR put
[#-*- coding:utf-8 -*-]
(except [ ] at the edge)
or something like that in the first line to make python understand Korean or your language or etc.
then it will work even if there is some Koreans in the path in my case.
So the things is, it seems like there is something about path or the letter.
People who answered are saying similar things. Hope you guys solve it!
I had the same problem and it turned out that my image names included special characters (e.g. château.jpg), which could not bet handled by cv2.imread. My solution was to make a temporary copy of the file, renaming it e.g. temp.jpg, which could be loaded by cv2.imread without any problems.
Note: I did not check the performance of shutil.copy2 vice versa other options. So probably there is a better/faster solution to make a temporary copy.
import shutil, sys, os, dlib, glob, cv2
for f in glob.glob(os.path.join(myfolder_path, "*.jpg")):
shutil.copy2(f, myfolder_path + 'temp.jpg')
img = cv2.imread(myfolder_path + 'temp.jpg')
img = cv2.cvtColor(img, cv2.COLOR_BGR2RGB)
os.remove(myfolder_path + 'temp.jpg')
If there are only few files with special characters, renaming can also be done as an exeption, e.g.
for f in glob.glob(os.path.join(myfolder_path, "*.jpg")):
try:
img = cv2.imread(f)
img = cv2.cvtColor(img, cv2.COLOR_BGR2RGB)
except:
shutil.copy2(f, myfolder_path + 'temp.jpg')
img = cv2.imread(myfolder_path + 'temp.jpg')
img = cv2.cvtColor(img, cv2.COLOR_BGR2RGB)
os.remove(myfolder_path + 'temp.jpg')
In my case it was a permission issue. I had to:
chmod a+wrx the image,
then it worked.
must please see guys that the error is in the cv2.imread() .Give the right path of the image. and firstly, see if your system loads the image or not. this can be checked first by simple load of image using cv2.imread().
after that ,see this code for the face detection
import numpy as np
import cv2
cascPath = "/Users/mayurgupta/opt/anaconda3/lib/python3.7/site- packages/cv2/data/haarcascade_frontalface_default.xml"
eyePath = "/Users/mayurgupta/opt/anaconda3/lib/python3.7/site-packages/cv2/data/haarcascade_eye.xml"
smilePath = "/Users/mayurgupta/opt/anaconda3/lib/python3.7/site-packages/cv2/data/haarcascade_smile.xml"
face_cascade = cv2.CascadeClassifier(cascPath)
eye_cascade = cv2.CascadeClassifier(eyePath)
smile_cascade = cv2.CascadeClassifier(smilePath)
img = cv2.imread('WhatsApp Image 2020-04-04 at 8.43.18 PM.jpeg')
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
faces = face_cascade.detectMultiScale(gray, 1.3, 5)
for (x,y,w,h) in faces:
img = cv2.rectangle(img,(x,y),(x+w,y+h),(255,0,0),2)
roi_gray = gray[y:y+h, x:x+w]
roi_color = img[y:y+h, x:x+w]
eyes = eye_cascade.detectMultiScale(roi_gray)
for (ex,ey,ew,eh) in eyes:
cv2.rectangle(roi_color,(ex,ey),(ex+ew,ey+eh),(0,255,0),2)
cv2.imshow('img',img)
cv2.waitKey(0)
cv2.destroyAllWindows()
Here, cascPath ,eyePath ,smilePath should have the right actual path that's picked up from lib/python3.7/site-packages/cv2/data here this path should be to picked up the haarcascade files
Your code can't find the figure or the name of your figure named the by error message.
Solution:
import cv2
import numpy as np
import matplotlib.pyplot as plt
img=cv2.imread('哈哈.jpg')#solution:img=cv2.imread('haha.jpg')
print(img)
If anyone is experiencing this same problem when reading a frame from a webcam [with code similar to "frame = cv2.VideoCapture(0)"] and work in Jupyter Notebook, you may try:
ensure previously tried code is not running already and restart Jupyter Notebook kernel
SEPARATE code "frame = cv2.VideoCapture(0)" in separate cell on place where it is [previous code put in cell above, code under put to cell down]
then run all the code above cell where is "frame = cv2.VideoCapture(0)"
then try run next cell with its only code "frame = cv2.VideoCapture(0)" - AND - till you will continue in executing other cells - ENSURE - that ASTERIX on the left side of this particular cell DISAPEAR and command order number appear instead - only then continue
now you can try execute the rest of your code as your camera input should not be empty anymore :-)
After end, ensure you close all your program and restart kernel to prepare it for another run
As #shaked litbak , this error arised with my initial use with the ASCII-generator , as i naively thought i just had to add to the ./data directory , with its load automatically .
I had to append the --input option with the desired file path .
I checked my image file path and it was correct. I made sure there was no corrupt images.The problem was with my mac. It sometimes have a hidden file called .DS_Store which was saved together with the image file path. Therefore cv2 was having a problem with that file.So I solved the problem by deleting .DS_Store
I also encountered this type of error:
error: OpenCV(4.1.2) /io/opencv/modules/imgproc/src/color.cpp:182: error: (-215:Assertion failed) !_src.empty() in function 'cvtColor'
The solution was to load the image properly. Since the file mentioned was wrong, images were not loaded and hence it threw this error. You can check the path of the image or if uploading an image through colab or drive, make sure that the image is present in the drive.
I encounter the problem when I try to load the image from non-ASCII path.
If I simply use imread to load the image, I am only able to get None.
Here is my solution:
import cv2
import numpy as np
path = r'D:\map\上海地图\abc.png'
image = cv2.imdecode(np.fromfile(path, dtype=np.uint8), cv2.IMREAD_UNCHANGED)
Similar thing will happen when I save the image in a non-ASCII path. It will not be successfully saved without any warnings. And here is what I did.
import cv2
import numpy as np
path = r'D:\map\上海地图\abc.png'
cv2.imencode('.png', image)[1].tofile(path)
path = os.path.join(raw_folder, folder, file)
print('[DEBUG] path:', path)
img = cv2.imread(path) #read path Image
if img is None: # check if the image exists in the path you give
print('Wrong path:', path)
else: # It completes the steps
img = cv2.resize(img, dsize=(128,128))
pixels.append(img)
The solution os to ad './' before the name of image before reading it...
Just Try Degrading the OpenCV
in python Shell (in cmd)
>>> import cv2
>>> cv2.__version__
after Checking in cmd
pip uninstall opencv-python
after uninstalling the version of opencv install
pip install opencv-python==3.4.8.29
Why would Pycharm put the second half of my dataset on a new line? (see Image) How do I turn this off?
I would like to display my dataset as wide as possible, with no wrapping.
The console attempts to auto-detect the width of the display area, but when that fails it defaults to 80 characters. This behavior can be overridden with:
import pandas as pd
pd.set_option('display.width', 400)
pd.set_option('display.max_columns', 10)
please take reference from this stackoverflow
Getting wider output in PyCharm's built-in console
I am trying to make a Choropleth map using matplotlib and cartopy for which I obviously need to plot a shapefile first. However, I did not manage to do so, even though a similar question has been asked here and here. I suspect either the projection or the bounds to be misspecified.
My shapefile has the projection
PROJCS["WGS_1984_UTM_Zone_32Nz",
GEOGCS["GCS_WGS_1984",
DATUM["WGS_1984",
SPHEROID["WGS_84",6378137,298.257223563]],
PRIMEM["Greenwich",0],
UNIT["Degree",0.017453292519943295]],
PROJECTION["Transverse_Mercator"],
PARAMETER["False_Easting",32500000],
PARAMETER["False_Northing",0],
PARAMETER["Central_Meridian",9],
PARAMETER["Scale_Factor",0.9996],
PARAMETER["Latitude_Of_Origin",0],
UNIT["Meter",1]]
and can be downloaded here where I am talking about vg250_2010-01-01.utm32w.shape.ebenen/vg250_ebenen-historisch/de1001/vg250_gem.shp
My code is
#!/usr/local/bin/python
# -*- coding: utf8 -*-
import cartopy.crs as ccrs
import cartopy.io.shapereader as shpreader
import matplotlib.pyplot as plt
fname = 'path/vg250_gem.shp'
proj = ccrs.TransverseMercator(central_longitude=0.0,central_latitude=0.0,
false_easting=32500000.0,false_northing=0.0,
scale_factor=0.9996)
municipalities = list(shpreader.Reader(fname).geometries())
ax = plt.axes(projection=proj)
plt.title('Deutschland')
ax.add_geometries(municipalities,proj,edgecolor='black',facecolor='gray',alpha=0.5)
ax.set_extent([32458044.649189778*0.9, 5556418.748046352*1.1, 32465287.307457082*0.9, 5564153.5456742775*1.1],proj)
plt.show()
where I obtained the bounds using the corresponding method from fiona. Python throws an error
Traceback (most recent call last):
File "***/src/analysis/test.py", line 16, in <module>
ax.set_extent([32458044.649189778, 5556418.748046352, 32465287.307457082, 5564153.5456742775],proj)
File "/usr/local/lib/python2.7/site-packages/cartopy/mpl/geoaxes.py", line 652, in set_extent
ylim=self.projection.y_limits))
ValueError: Failed to determine the required bounds in projection coordinates. Check that the values provided are within the valid range (x_limits=[-20000000.0, 20000000.0], y_limits=[-10000000.0, 10000000.0]).
[Finished in 53.9s with exit code 1]
This doesn't make sense to me. Also, experimenting with ccrs.UTM() gives me a plot showing a white area. I'd appreciate it if anyone can tell me how to fix this. Thank you!
I have found two issues. One is an incorrect specification of limits in your call to set_extent, the documentation specifies [x0,x1,y0,y1] should be the input, you seem to have given [x0,y0,x1,y1].
The other issue seems to be a limitation in cartopy, as best I can tell. It looks like projections outside the limits listed in the error message will always fail, and those limits are hardcoded. You can just edit the source (this line in their latest release), changing -2e7 to -4e7, likewise for the upper bound. After these fixes, your plot is generated without issue:
The new set_extent line:
ax.set_extent([32458044.649189778*0.975, 32465287.307457082*1.025,5556418.748046352*0.9, 556415,3.5456742775*1.1],proj)
You may also want to set central_longitude=9.0 in your TransverseMercator, that seems to be what's specified in your shapefile.
I would recommend contacting the developers about this, they might have a good reason for setting those bounds, or they might have a better workaround, or maybe they'll widen the bounds in a later release!
Update
Your bounds also seem to have been set based on only the first of the municipalities:
In [34]: municipalities[0].bounds
Out[34]: (32458044.649189778, 5556418.748046352, 32465287.307457082, 5564153.5456742775)
But the other elements have different bounds. You can get limits flushed to the actual drawing based on min/max values of the bounds of all municipalities.
bnd = np.array([i.bounds for i in municipalities])
x0,x1 = np.min(bnd[:,0]),np.max(bnd[:,2])
y0,y1 = np.min(bnd[:,1]),np.max(bnd[:,3])
ax.set_extent([x0,x1,y0,y1],proj)
I am having problems saving a figure created by matplotlib as a .eps or .ps when I enable usetex=True. This works when this is not enabled. Here's an example:
plt.plot([1,2,3], [1,2,3], 'b.')
plt.text(2,2,r'\textbf{(a)} \lambda_{1} value', usetex=True, fontsize=16, fontname='Times New Roman')
plt.savefig('check.eps')
I receive this error:
File "C:\Python27\lib\site-packages\matplotlib\backends\backend_ps.py", line 671, in draw_tex
thetext = 'psmarker%d' % self.textcnt
AttributeError: 'RendererPS' object has no attribute 'textcnt'
I also cannot set the font to Time New Roman using the text command when I enable usetex=True.
Even if it is a bit late, I would like to give an answer, if someone else searches for this error.
Saving eps files starts the RendererPS backend, which checks whether plt.rcParams['text.usetex'] is set to True to initialise its textcnt attribute. Since usetex is set only for the text object, the backend does not expect to have to handle LaTeX and throws an error when it tries to. These kind of inconsistencies concerning LaTeX rendering are sadly present in a number of places in matplotlib.
If you cannot achieve the text formatting with the standard matplotib functionality, one solution is to set usetex globally: plt.rcParams['text.usetex'] = True. This render all the text of the figure with LaTeX (e.g. also tick labels or axis labels). I would recommend to do this in any case, to have consistent visuals.
Concerning the font, the fontname argument only affects standard matplotlib formatting. For LaTeX rendering, you have to specify the font like you would do in LaTeX. To get a font like Times New Roman, you need to load the respective package e.g. mathptmx (the times package is deprecated) by plt.rcParams['text.latex.preamble'] = [r'\usepackage{mathptmx}']. Of course, the package has to be installed in your local LaTeX installation. The standard font family in matplotlib is sans-serif, so this has to be changed to serif with plt.rcParams['font.family'] = 'serif'.
The final code is:
import matplotlib.pyplot as plt
# LaTeX setup
plt.rcParams['text.latex.preamble'] = [r'\usepackage{mathptmx}'] # load times roman font
plt.rcParams['font.family'] = 'serif' # use serif font as default
plt.rcParams['text.usetex'] = True # enable LaTeX rendering globally
plt.plot([1,2,3], [1,2,3], 'b.')
plt.text(2,2,r'\textbf{(a)} $\lambda_1$ value', fontsize=16)
plt.savefig('check.eps')
Note, that you have to enclose the math variable in dollar signs $\lambda_1$, otherwise you get an error or warning in LaTeX. (Also, single character subscripts do not need to be enclosed in curly brackets).
As a side note: I encountered the same error, when trying to save a figure as eps after turning usetex off again.
I wrote a script that calls functions from QIIME to build a bunch of plots among other things. Everything runs fine to completion, but matplotlib always throws the following feedback for every plot it creates (super annoying):
/usr/local/lib/python2.7/dist-packages/matplotlib/pyplot.py:412: RuntimeWarning: More than 20 figures have been opened. Figures created through the pyplot interface (matplotlib.pyplot.figure) are retained until explicitly closed and may consume too much memory. (To control this warning, see the rcParam figure.max_num_figures).
max_open_warning, RuntimeWarning)
I found this page which seems to explain how to fix this problem , but after I follow directions, nothing changes:
import matplotlib as mpl
mpl.rcParams[figure.max_open_warning'] = 0
I went into the file after calling matplotlib directly from python to see which rcparams file I should be investigating and manually changed the 20 to 0. Still no change. In case the documentation was incorrect, I also changed it to 1000, and still am getting the same warning messages.
I understand that this could be a problem for people running on computers with limited power, but that isn't a problem in my case. How can I make this feedback go away permanently?
Try setting it this way:
import matplotlib as plt
plt.rcParams.update({'figure.max_open_warning': 0})
Not sure exactly why this works, but it mirrors the way I have changed the font size in the past and seems to fix the warnings for me.
Another way I just tried and it worked:
import matplotlib as mpl
mpl.rc('figure', max_open_warning = 0)
When using Seaborn you can do it like this
import seaborn as sns
sns.set_theme(rc={'figure.max_open_warning': 0})
Check out this article which basically says to plt.close(fig1) after you're done with fig1. This way you don't have too many figs floating around in memory.
In Matplotlib, figure.max_open_warning is a configuration parameter that determines the maximum number of figures that can be opened before a warning is issued. By default, the value of this parameter is 20. This means that if you open more than 20 figures in a single Matplotlib session, you will see a warning message. You can change the value of this parameter by using the matplotlib.rcParams function. For example:
import matplotlib.pyplot as plt
plt.rcParams['figure.max_open_warning'] = 50
This will set the value of figure.max_open_warning to 50, so that you will see a warning message if you open more than 50 figures in a single Matplotlib session.