Biologists, sometimes are involved in the preparation of field guides for species identification. Arrange many images could be an endless task. Because I am lazy, I always try to automatize the work.

Recently, I had to arrange 60 images in ten arrays of 2x3 (2 columns, three rows). Luckily, I found a code in Python that can do the work for me. The code was written for Evan Rosica, and it can be found here.

Herein, an example of how the code works!

We are going to merge the following images in a matrix of 2x2:

In #1 I am loading the libraries required. Basically, this code use "PIL."

In #2 you need to specify the directory where your images are stored (line 16).

In #3 the most relevant is that you should understand that all images will be resized according to the first image of your directory (line 28). However, you can change it quickly.

In #4 is a loop which will add each image in a sequence. There is not much mystery here.

In #5 I create some boxes and label for my images. I think it is possible to create a loop for this. If you can improve it you are welcome to try!

In #6 The final result is showed and saved if you want (uncomment line 84).

Below the full code

And here, the final result.

As you can see this code can help us to save time if we need to arrange too many images, even if you are writing papers that demand merging a lot of figures.

# Biological Complexity

Methods for studying biological phenomena.

## Thursday, 14 February 2019

## Tuesday, 14 August 2018

### Categorize a Raster (DEM)

Although Raster files are visually attractive, they provide hard to interpret information because of is representing continuous data. So, they often are discretized in categories. There are many procedures to get it done (try QGis), but on this occasion, I will show how to do in R.

First, we need to load two libraries rgdal and raster, if you are using ess in emacs perhaps you must do some additional steps (you can email me). Now, we load our raster file, here, I'll use a Digital Elevation Model (DEM) of my country Peru (DEM_PERÚ.tif). You can download all the files in this link. However, this raster file doesn't match with the real form of my country, so we'll need to crop it, we'll do it later, but we are going to take advantage and load a vectorial shapefile which will be our mask (peru_disolved.shp). I recommend looking at the graphics to be sure that our files are loaded correctly.

The graphic is weird because we still need to crop it as a sixth step. You can easily get it using the following code.

If you wish, you can save the files:

All the code is:

That's all for today.

First, we need to load two libraries rgdal and raster, if you are using ess in emacs perhaps you must do some additional steps (you can email me). Now, we load our raster file, here, I'll use a Digital Elevation Model (DEM) of my country Peru (DEM_PERÚ.tif). You can download all the files in this link. However, this raster file doesn't match with the real form of my country, so we'll need to crop it, we'll do it later, but we are going to take advantage and load a vectorial shapefile which will be our mask (peru_disolved.shp). I recommend looking at the graphics to be sure that our files are loaded correctly.

```
# 1. Packages
library(rgdal) # Geospatial Data Abstraction Library
library(raster) # Tools for gridded spatial data
# 2. Read DEM and convert to raster layer object
dem <- raster("~/Downloads/DEM_PERÚ.tif") # raster
vperu <- readOGR("peru_disolved.shp", layer="peru_disolved") # vectorial
plot(dem) # preliminar plotting
```

Third, we are going to define the levels which will represent the categories. I have chosen only six levels (you can try others). We save the levels as a matrix. Fourth, using the function reclassify, we'll obtain a new file with the only the levels previously defined. Fifth, we obtain the graphic of the reclassified DEM.```
# 3. Generate a reclass matrix
m <- c(-Inf, 250, 1, 250, 500, 2, 500, 1000, 3, 1000, 2000, 4,
2000, 4500, 5, 4500, Inf, 6) # specify dsired levels
rclmat <- matrix(m, ncol = 3, byrow = TRUE)
# 4. Reclassify the raster layer
reclassified_dem <- reclassify(dem, rclmat)
# 5. Plot the results
par(mfrow=c(1, 2))
plot(dem)
plot(reclassified_dem)
```

The graphic is weird because we still need to crop it as a sixth step. You can easily get it using the following code.

```
# 6. Cropping
r1 <- crop(reclassified_dem, extent(vperu))
r2 <- mask(r1, vperu)
plot(r2)
```

If you wish, you can save the files:

```
# 7. Save new raster
writeRaster(reclassified_dem, "~/Downloads/reclassified_dem.tif", drivername="GTiff",
type = "Float32")
writeRaster(dem_crop, "~/Downloads/dem_crop.tif", drivername="GTiff",
type = "Float32")
```

All the code is:

```
# -*- coding: utf-8 -*-
# Author: Irbin B. Llanqui
#"""
# Language: R script
# This is a temporary script file.
#"""
# 1. Packages
library(rgdal) # Geospatial Data Abstraction Library
library(raster) # Tools for gridded spatial data
# 2. Read DEM and convert to raster layer object
dem <- raster("~/Downloads/DEM_PERÚ.tif") # raster
vperu <- readOGR("peru_disolved.shp", layer="peru_disolved") # vectorial
plot(dem) # preliminar plotting
# 3. Generate a reclass matrix
m <- c(-Inf, 250, 1, 250, 500, 2, 500, 1000, 3, 1000, 2000, 4,
2000, 4500, 5, 4500, Inf, 6) # specify dsired levels
rclmat <- matrix(m, ncol = 3, byrow = TRUE)
# 4. Reclassify the raster layer
reclassified_dem <- reclassify(dem, rclmat)
# 5. Plot the results
par(mfrow=c(1, 2))
plot(dem)
plot(reclassified_dem)
# 6. Cropping
r1 <- crop(reclassified_dem, extent(vperu))
r2 <- mask(r1, vperu)
plot(r2)
# 7. Save new raster
writeRaster(reclassified_dem, "~/Downloads/reclassified_dem.tif", drivername="GTiff",
type = "Float32")
writeRaster(dem_crop, "~/Downloads/dem_crop.tif", drivername="GTiff",
type = "Float32")
```

That's all for today.

## Monday, 5 March 2018

### A common error in adehabitatHR

Recently I figured out that in some forum people are asking about this error in the library adehabitatHR of R:

Error in getverticeshr.estUD(x[[i]], percent, ida = names(x)[i], unin,: The grid is too small to allow the estimation of home-range. You should rerun kernelUD with a larger extent parameter

Well, I'll try to explain this and solve it. First of all, this error occurs in the workflow to obtain polygons from a KDE volume, which is a common procedure in home range analysis.

In the first four steps of the following code, I simulate a dataset, which will represent the spatial position of two individuals.

Now, I will estimate the kernel density using those spatial points.

And we obtain these nice plots. Now, we want to extract the contour lines at 75% of probability density. For that we will use the function get vertices as the following code:

And we will get an ERROR!! specifically, we will obtain the error I introduced at the beginning of this post.

But, why this happened? If you are a shrewd observer, you'll notice in the above plots that, the contour line at 75% is cut, and our domain doesn't include it. So, that is the reason for the error, is that R can't estimate the vertices of the contour line at 75% precisely because they are no in the domain, they were not computed. On contrary, the contour line at 50% is perfectly inside the domain, so if we ask for this vertices we won't have any error.

Now, if we want to extract 75% contour lines without an error, we only need to increase the grid in order to cover all the contour lines. In this case, I will increase the grid at 50 (see the item # 5. Domain x <- seq(-50, 150, by=1.) y <- seq(-50, 150, by=1.))

Now, all the contour lines are inside the grid, so we'll no see the message of error. And that is.

Error in getverticeshr.estUD(x[[i]], percent, ida = names(x)[i], unin,: The grid is too small to allow the estimation of home-range. You should rerun kernelUD with a larger extent parameter

Well, I'll try to explain this and solve it. First of all, this error occurs in the workflow to obtain polygons from a KDE volume, which is a common procedure in home range analysis.

In the first four steps of the following code, I simulate a dataset, which will represent the spatial position of two individuals.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 | # -*- coding: utf-8 -*- # Author: Irbin B. Llanqui #""" # Language: R script # This is a temporary script file. #""" # 1. Packages library(adehabitatHR) # Package for spatal analysis # 2. Empty Dataframe points <- data.frame(ID = double()) XY_cor <- data.frame(X = double(), Y = double()) # 3. Assigning values (this will be our spatial coordinates) set.seed(17) for(i in c(1:100)){ if(i >= 50){points[i, 1] <- 1} else {points[i, 1] <- 2} XY_cor[i, 1] <- runif(1, 0, 100) XY_cor[i, 2] <- runif(1, 0, 100)} # 4. Transform to SpatialDataframe coordinates(points) <- XY_cor[, c("X", "Y")] class(points) |

Now, I will estimate the kernel density using those spatial points.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 | # 5. Domain x <- seq(0, 100, by=1.) # resolution is the pixel size you desire y <- seq(0, 100, by=1.) xy <- expand.grid(x=x,y=y) coordinates(xy) <- ~x+y gridded(xy) <- TRUE class(xy) # 6. Kernel Density kud_points <- kernelUD(points, h = "href", grid = xy) image(kud_points) # 7. Get the Volum vud_points <- getvolumeUD(kud_points) # 8. Get contour levels <- c(50, 75, 95) list <- vector(mode="list", length = 2) list[[1]] <- as.image.SpatialGridDataFrame(vud_points[[1]]) list[[2]] <- as.image.SpatialGridDataFrame(vud_points[[2]]) # 9. Plot par(mfrow = c(2, 1)) image(vud_points[[1]]) contour(list[[1]], add=TRUE, levels=levels) image(vud_points[[2]]) contour(list[[2]], add=TRUE, levels=levels) |

And we obtain these nice plots. Now, we want to extract the contour lines at 75% of probability density. For that we will use the function get vertices as the following code:

1 2 3 4 | # 10. Get vertices (Will be an Error) vkde_points <- getverticeshr(kud_points, percent = 75, unin = 'm', unout='m2') plot(vkde_points) |

And we will get an ERROR!! specifically, we will obtain the error I introduced at the beginning of this post.

**Error in getverticeshr.estUD(x[[i]], percent, ida = names(x)[i], unin,: The grid is too small to allow the estimation of home-range. You should rerun kernelUD with a larger extent parameter**But, why this happened? If you are a shrewd observer, you'll notice in the above plots that, the contour line at 75% is cut, and our domain doesn't include it. So, that is the reason for the error, is that R can't estimate the vertices of the contour line at 75% precisely because they are no in the domain, they were not computed. On contrary, the contour line at 50% is perfectly inside the domain, so if we ask for this vertices we won't have any error.

1 2 3 4 | # 10. Get vertices (Will be an Error) vkde_points <- getverticeshr(kud_points, percent = 50, unin = 'm', unout='m2') plot(vkde_points) |

Now, if we want to extract 75% contour lines without an error, we only need to increase the grid in order to cover all the contour lines. In this case, I will increase the grid at 50 (see the item # 5. Domain x <- seq(-50, 150, by=1.) y <- seq(-50, 150, by=1.))

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 | # 5. Domain x <- seq(-50, 150, by=1.) # resolution is the pixel size you desire y <- seq(-50, 150, by=1.) xy <- expand.grid(x=x,y=y) coordinates(xy) <- ~x+y gridded(xy) <- TRUE class(xy) # 6. Kernel Density kud_points <- kernelUD(points, h = "href", grid = xy) image(kud_points) # 7. Get the Volum vud_points <- getvolumeUD(kud_points) # 8. Get contour levels <- c(50, 75, 95) list <- vector(mode="list", length = 2) list[[1]] <- as.image.SpatialGridDataFrame(vud_points[[1]]) list[[2]] <- as.image.SpatialGridDataFrame(vud_points[[2]]) # 9. Plot par(mfrow = c(2, 1)) image(vud_points[[1]]) contour(list[[1]], add=TRUE, levels=levels) image(vud_points[[2]]) contour(list[[2]], add=TRUE, levels=levels) # 10. Get vertices (Will be an Error) vkde_points <- getverticeshr(kud_points, percent = 75, unin = 'm', unout='m2') plot(vkde_points) |

Now, all the contour lines are inside the grid, so we'll no see the message of error. And that is.

## Sunday, 28 January 2018

### On species accumulation curves

On this occasion, I will share some quick tips related to the implementation of Species Accumulation Curves (SAC) using R (in another occasion I'll use Python of course). First of all, we have to remember that SAC is a tool to assess and/or compare the alpha diversity. For instance, we can imagine that someone who wants to know the alpha diversity (Species richness) in a certain place, for that, a sampling is performed. If we imagine this scenario we can realize that while the sampling is performed the number of new species registered will decrease respect the time. That is because, first in the sampling, the common species will be detected, but later, will be harder find new species because only remain rare. This can be visualized in the form of the below curve (Moreno & Haffer, 2000):

As you can see, the curve increase with few sampling units, however, as long as the sampling continue the curve increase even slower. When the increase of new species tend to zero (asymptote) that mean that the inventory is complete, therefore that is the richness of the study area. In R, there some way to obtain this accumulation curves using field data, the most common are using "vegan" and "BiodiversityR" packages:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 | # Dependences library("BiodiversityR") library("vegan") # Loading data data(BCI) # Using vegan package sp1 <- specaccum(BCI, method = "exact") sp2 <- specaccum(BCI, method = "random") # Using BiodiversityR package sp3 <- accumresult(BCI, method = "exact") sp4 <- accumresult(BCI, method = "random") # Comparing results using plots par(mfrow=c(1,2)) plot(sp1, col = "black", lwd = 3, xlab = "Samples", ylab = "Richness", main = "exact") plot(sp3, col = "red", xlab = "Samples", ylab = "Richness", add = TRUE) plot(sp2, col = "black", lwd = 3, xlab = "Samples", ylab = "Richness", main = "random") plot(sp4, col = "red", xlab = "Samples", ylab = "Richness", add = TRUE) legend("bottomright", c("vegan","BiodiversityR"), col = c("black","red"), lty = 1) |

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 | # Dependences library("BiodiversityR") library("vegan") # Loading data data(BCI) # Using vegan package sp1 <- specaccum(BCI, method = "exact") sp2 <- specaccum(BCI, method = "random") # Using BiodiversityR package sp3 <- accumresult(BCI, method = "exact") sp4 <- accumresult(BCI, method = "random") # Comparing results using plots par(mfrow=c(1,2)) plot(sp1, col = "black", lwd = 3, xlab = "Samples", ylab = "Richness", main = "exact") plot(sp3, col = "red", xlab = "Samples", ylab = "Richness", add = TRUE) plot(sp2, col = "black", lwd = 3, xlab = "Samples", ylab = "Richness", main = "random") plot(sp4, col = "red", xlab = "Samples", ylab = "Richness", add = TRUE) legend("bottomright", c("vegan","BiodiversityR"), col = c("black","red"), lty = 1) # Estimators sp1_pool <- poolaccum(BCI, permutations = 1000) plot(sp1_pool) |

The object created by "poolaccum" function by default give a plot using parameters of the "lattice" package:

But, it is not a clean plot if we want to use the plot for a research presentation. For example, likely we want to obtain just one plot without the frame, and all in black. Fortunately, we can modify the plot at our convenience. For example, Chao estimator use to be a great way to assess the completeness of an inventory, so we want just this plot in an elegant way, so we need to delete the other plots, remove some axes, box titles, and change the colour, for this we can use the following code:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 | # Dependences library("BiodiversityR") library("vegan") # Loading data data(BCI) # Using vegan package sp1 <- specaccum(BCI, method = "exact") sp2 <- specaccum(BCI, method = "random") # Using BiodiversityR package sp3 <- accumresult(BCI, method = "exact") sp4 <- accumresult(BCI, method = "random") # Comparing results using plots par(mfrow=c(1,2)) plot(sp1, col = "black", lwd = 3, xlab = "Samples", ylab = "Richness", main = "exact") plot(sp3, col = "red", xlab = "Samples", ylab = "Richness", add = TRUE) plot(sp2, col = "black", lwd = 3, xlab = "Samples", ylab = "Richness", main = "random") plot(sp4, col = "red", xlab = "Samples", ylab = "Richness", add = TRUE) legend("bottomright", c("vegan","BiodiversityR"), col = c("black","red"), lty = 1) # Estimators sp1_pool <- poolaccum(BCI, permutations = 1000) plot(sp1_pool) # Plot manipulating lattice plot(sp1_pool, display = "chao", col = "black", auto.key = FALSE, grid = F, strip = FALSE, xlab = "Sample", par.settings = list(axis.line = list(col = 0)), scales = list(col=1, tck=c(1,0)), panel = function(...){ lims <- current.panel.limits() panel.xyplot(...) panel.abline(h=lims$ylim[1], v=lims$xlim[1]) }) |

Done!, so we have a predicted accumulation curve with 95% confidence intervals (CI). A quick look, show us that the inventory is still incomplete because the CI of the Chao estimator is not stable, so in the next plot we can see another plot where more sampling units have been added and how the Chao estimator get stable, and therefore, in this case, we can conclude that the inventory is complete and the final richness or number of species is equal to the value of the asymptote of the Chao estimator.

**REFERENCES**

Chiarucci, A., Bacaro, G., Rocchini, D., & Fattorini, L. (2008). Discovering and rediscovering the sample-based rarefaction formula in the ecological literature.

*Community Ecology*,*9*(1), 121–123. http://doi.org/10.1556/ComEc.9.2008.1.14
Gotelli, N. J., & Colwell, R. K. (2001). Quantifying biodiversity: Procedures and pitfalls in the measurement and comparison of species richness.

*Ecology Letters*,*4*(4), 379–391. http://doi.org/10.1046/j.1461-0248.2001.00230.x
Moreno, C. E., & Halffter, G. (2000). Assessing the completeness of bat biodiversity inventories using species accumulation curves.

*Journal of Applied Ecology*,*37*(1), 149–158. http://doi.org/10.1046/j.1365-2664.2000.00483.x## Monday, 24 July 2017

### Projecting coordinates in Python

Sometimes we have to manipulate huge spatial databases and use to be a headband if we must to convert the spatial point to a different projection. Well, a nice way to solve this sort of issues is take advantage of the libraries of python, in this case I am going to describe how to use the package utm for it.

First, we are going to simulate some spatial data in a decimal projection.

Done!, we have already created a data frame with a hundred of spatial points in decimal degrees. now, we are going to project them to UTM. First we create a temporal "temp" variable who store our XY coordinates projected. After that, ee are going to use the function utm.from_latlon() in a loop. This one put each project coordinate in your recently created variable "temp"

Finally, we convert the data project to a DataFrame and assign this to a new variable "data", and that all. In fact, this process help me to project a database with two hundred thousand locations, so is useful to project huge databases.

First, we are going to simulate some spatial data in a decimal projection.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 | import matplotlib.pyplot as plt import numpy as np import pandas as pd import utm lat = np.linspace(20, 40, 100) long = np.linspace(60, 70, 100) index = np.arange(1,101) d = {'x': pd.Series(lat, index=index), 'y': pd.Series(long, index=index)} df = pd.DataFrame(d) print(df) |

Done!, we have already created a data frame with a hundred of spatial points in decimal degrees. now, we are going to project them to UTM. First we create a temporal "temp" variable who store our XY coordinates projected. After that, ee are going to use the function utm.from_latlon() in a loop. This one put each project coordinate in your recently created variable "temp"

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 | import matplotlib.pyplot as plt import numpy as np import pandas as pd import utm lat = np.linspace(20, 40, 100) long = np.linspace(60, 70, 100) index = np.arange(1,101) d = {'x': pd.Series(lat, index=index), 'y': pd.Series(long, index=index)} df = pd.DataFrame(d) #print(df) temp = [[None]*1 for ii in range(len(df.x))] ## Use the utm package to transform coordinates for i in np.arange(0, len(df.x)): temp[i] = utm.from_latlon(df.y[i+1], df.x[i+1]) # Transform in a dataframe data = pd.DataFrame(temp) data.columns = ['X', 'Y', 'Zone', 'Code'] print(data) |

Finally, we convert the data project to a DataFrame and assign this to a new variable "data", and that all. In fact, this process help me to project a database with two hundred thousand locations, so is useful to project huge databases.

## Thursday, 20 July 2017

### Diffusion equation

Diffusion equation is widely discussed in the literature, specially in physics and mathematics. However, diffusion equation also has relevance in other fields, for instance biology, where it can be really useful to understand some biological phenomena at different level, including population biology. Therefore, is important to know how to solve it.

Diffusion equation is a partial differential equation which involve both a time and space. In 1D it would be:

$\frac{\partial u}{\partial t} = D \frac{\partial^2 u}{\partial x^2}$

it can be discretised using numerical methods. In this case we are going to use the finite difference method, whereby the above equation can be discretised as:

$\frac{u^{t+1}_{i} - u^{t+1}_{i}}{\Delta t} = D \frac{u^t_{i+1} - 2u^t_{i} + u^t_{i-1}}{\Delta^2x}$

where, $t$ indicate the time set and $i$ the x position.

For obtaining the steady state we are going to programming this in python. The below code solve numerically the diffusion equation

The solution is

Where is showed how the distribution is diffused along the domain. However, as time goes by, the distribution density tend to zero, whereby the mass conservative property is not satisfied. Whether we can maintain a constant mass, we need to apply boundary conditions, in this case a zero flux in the extremes of the domain. For that, we establish:

$\frac{\partial u}{ \partial t} = 0 $

Here, zero flux condition is included straightforward adding u[0] = u[1] (left) and

u[-1]= u[-2] (right). Then, the mass can not go out the domain.

But!, be careful, because this zero flux condition only works for this simple equation. If you are working with other ones, for instance advection-difussion, zero flux condition has to be implemented in a different way.

Diffusion equation is a partial differential equation which involve both a time and space. In 1D it would be:

$\frac{\partial u}{\partial t} = D \frac{\partial^2 u}{\partial x^2}$

it can be discretised using numerical methods. In this case we are going to use the finite difference method, whereby the above equation can be discretised as:

$\frac{u^{t+1}_{i} - u^{t+1}_{i}}{\Delta t} = D \frac{u^t_{i+1} - 2u^t_{i} + u^t_{i-1}}{\Delta^2x}$

where, $t$ indicate the time set and $i$ the x position.

For obtaining the steady state we are going to programming this in python. The below code solve numerically the diffusion equation

import numpy as np import matplotlib.pyplot as plt #Setting parameters nx = 101 dx = 0.01 nt = 20000 D = .1 dt = 0.0001 #be aware setting this value #Initial conditions u = np.zeros(nx) u[25:75] = 2 un = u.copy() #Numerical scheme: Finite differences for n in range(nt): for i in range(1, nx-1): un = u.copy() u[i] = un[i] + D * dt /dx**2* (un[i+1]-2*un[i]+un[i-1]) #Plot the results plt.plot(np.linspace(0, 1, 101), u) plt.ylim(0, 4) plt.xlim(0, 1.) plt.show()

The solution is

Where is showed how the distribution is diffused along the domain. However, as time goes by, the distribution density tend to zero, whereby the mass conservative property is not satisfied. Whether we can maintain a constant mass, we need to apply boundary conditions, in this case a zero flux in the extremes of the domain. For that, we establish:

$\frac{\partial u}{ \partial t} = 0 $

import numpy as np import matplotlib.pyplot as plt #Setting parameters nx = 101 dx = 0.01 nt = 20000 D = .1 dt = 0.0001 #be aware setting this value #Initial conditions u = np.zeros(nx) u[25:75] = 2 un = u.copy() #Numerical scheme: Finite differences for n in range(nt): for i in range(1, nx-1): un = u.copy() u[i] = un[i] + D * dt /dx**2* (un[i+1]-2*un[i]+un[i-1]) # Zero flux condition u[0] = u[1] u[-1]= u[-2] #Plot the results plt.plot(np.linspace(0, 1, 101), u) plt.ylim(0, 4) plt.xlim(0, 1.) plt.show()

Here, zero flux condition is included straightforward adding u[0] = u[1] (left) and

u[-1]= u[-2] (right). Then, the mass can not go out the domain.

But!, be careful, because this zero flux condition only works for this simple equation. If you are working with other ones, for instance advection-difussion, zero flux condition has to be implemented in a different way.

## Sunday, 16 July 2017

### Updating spyder

If you are a python users, maybe you use spyder as your IDLE. For update your spyder version the common procedure is by the instruction:

conda update spyder

which should work if you have conda package manager. Moreover you can also upgrade spyder for specific environments, for instance, if we have a python environment called py27, we can upgrade spyder with:

conda update -n py27 spyder

However, whether the above does not work, another way is to use the pip instruction:

pip install --upgrade spyder

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