Iteratively generating subplots in matplotlib by row and column - only final axes plotting
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I am writing code to plot cross correlations of every time-series in my data against all others, with two for-loops to index the row and column position respectively (column loop nested within the row loop).
Currently only the final axes (i.e. bottom right corner) of the figure is displaying any data, and each iteration of the loop appears to be plotting on this axes. I am wondering if I have made any obvious mistakes with the order of commands in the nested for loops, or if I am misinterpreting the input arguments to matplotlib functions like subplots....
The code is as below:
fig, axes = plt.subplots(nrows=data_num, ncols=data_num, sharex=True, sharey=True)
for n in range(data_num): #row index
for p in range(data_num): # column index
x = data_df.iloc[:,n] #get data for ROI according to row index
print(x.head())
x = x.values
y = data_df.iloc[:,p] #get data for ROI according to column index
print(y.head())
y = y.values
axes[n,p] = plt.xcorr(x,y,normed=True) #axes [row,column] = cross correlation plot of above data
print(f'plotting at index [ {n} , {p}]')
python python-3.x matplotlib
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up vote
0
down vote
favorite
I am writing code to plot cross correlations of every time-series in my data against all others, with two for-loops to index the row and column position respectively (column loop nested within the row loop).
Currently only the final axes (i.e. bottom right corner) of the figure is displaying any data, and each iteration of the loop appears to be plotting on this axes. I am wondering if I have made any obvious mistakes with the order of commands in the nested for loops, or if I am misinterpreting the input arguments to matplotlib functions like subplots....
The code is as below:
fig, axes = plt.subplots(nrows=data_num, ncols=data_num, sharex=True, sharey=True)
for n in range(data_num): #row index
for p in range(data_num): # column index
x = data_df.iloc[:,n] #get data for ROI according to row index
print(x.head())
x = x.values
y = data_df.iloc[:,p] #get data for ROI according to column index
print(y.head())
y = y.values
axes[n,p] = plt.xcorr(x,y,normed=True) #axes [row,column] = cross correlation plot of above data
print(f'plotting at index [ {n} , {p}]')
python python-3.x matplotlib
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am writing code to plot cross correlations of every time-series in my data against all others, with two for-loops to index the row and column position respectively (column loop nested within the row loop).
Currently only the final axes (i.e. bottom right corner) of the figure is displaying any data, and each iteration of the loop appears to be plotting on this axes. I am wondering if I have made any obvious mistakes with the order of commands in the nested for loops, or if I am misinterpreting the input arguments to matplotlib functions like subplots....
The code is as below:
fig, axes = plt.subplots(nrows=data_num, ncols=data_num, sharex=True, sharey=True)
for n in range(data_num): #row index
for p in range(data_num): # column index
x = data_df.iloc[:,n] #get data for ROI according to row index
print(x.head())
x = x.values
y = data_df.iloc[:,p] #get data for ROI according to column index
print(y.head())
y = y.values
axes[n,p] = plt.xcorr(x,y,normed=True) #axes [row,column] = cross correlation plot of above data
print(f'plotting at index [ {n} , {p}]')
python python-3.x matplotlib
I am writing code to plot cross correlations of every time-series in my data against all others, with two for-loops to index the row and column position respectively (column loop nested within the row loop).
Currently only the final axes (i.e. bottom right corner) of the figure is displaying any data, and each iteration of the loop appears to be plotting on this axes. I am wondering if I have made any obvious mistakes with the order of commands in the nested for loops, or if I am misinterpreting the input arguments to matplotlib functions like subplots....
The code is as below:
fig, axes = plt.subplots(nrows=data_num, ncols=data_num, sharex=True, sharey=True)
for n in range(data_num): #row index
for p in range(data_num): # column index
x = data_df.iloc[:,n] #get data for ROI according to row index
print(x.head())
x = x.values
y = data_df.iloc[:,p] #get data for ROI according to column index
print(y.head())
y = y.values
axes[n,p] = plt.xcorr(x,y,normed=True) #axes [row,column] = cross correlation plot of above data
print(f'plotting at index [ {n} , {p}]')
python python-3.x matplotlib
python python-3.x matplotlib
asked Nov 20 at 12:38
OsDavy
83
83
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1 Answer
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It's a, perhaps unfortunate, way that pyplot and matplotlib works: you have to create the plots on the respective axes, not assign the result from a pyplot.xcorr
call to an axes. Thus: axis[n,p].xcorr(...)
. So the interface is suddenly somewhat more object-oriented than the usual direct pyplot
calls.
All the plots ends up in just the last figure, because you are calling
plt.xcorr(x,y,normed=True)
It doesn't matter if you then assign the return value to the axes array elements, which you shouldn't, as that destroys the original axes array.plt.xcorr
will then plot all the data in the same plot on top of each other, because pyplot generally acts on the currently active axes, which is the last one created via plt.subplots()
.
That's for an explanation. Here's an example solution (with random data and a simple scatter plot):
import numpy as np
import matplotlib.pyplot as plt
data_num = 3
x = np.random.uniform(1, 10, size=(data_num, data_num, 20))
y = np.random.uniform(5, 20, size=(data_num, data_num, 20))
fig, axes = plt.subplots(nrows=data_num, ncols=data_num, sharex=True, sharey=True)
for n in range(data_num): #row index
for p in range(data_num): # column index
# Call `scatter` or any plot function on the
# respective `axes` object itself
axes[n,p].scatter(x[n,p], y[n,p])
print(f'plotting at index [ {n} , {p}]')
plt.savefig('figure.png')
and figure.png looks like (sorry, no colour or symbol variation, just bare bones scatter plots):
Brilliant, it was literally a case of changing that one line from 'axes[n,p] = plt.xcorr .......' to 'axes[n,p].xcorr(....' Really helpful, thank you!
– OsDavy
Nov 20 at 13:08
I corrected the part about the explanation.
– ImportanceOfBeingErnest
Nov 20 at 13:56
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
It's a, perhaps unfortunate, way that pyplot and matplotlib works: you have to create the plots on the respective axes, not assign the result from a pyplot.xcorr
call to an axes. Thus: axis[n,p].xcorr(...)
. So the interface is suddenly somewhat more object-oriented than the usual direct pyplot
calls.
All the plots ends up in just the last figure, because you are calling
plt.xcorr(x,y,normed=True)
It doesn't matter if you then assign the return value to the axes array elements, which you shouldn't, as that destroys the original axes array.plt.xcorr
will then plot all the data in the same plot on top of each other, because pyplot generally acts on the currently active axes, which is the last one created via plt.subplots()
.
That's for an explanation. Here's an example solution (with random data and a simple scatter plot):
import numpy as np
import matplotlib.pyplot as plt
data_num = 3
x = np.random.uniform(1, 10, size=(data_num, data_num, 20))
y = np.random.uniform(5, 20, size=(data_num, data_num, 20))
fig, axes = plt.subplots(nrows=data_num, ncols=data_num, sharex=True, sharey=True)
for n in range(data_num): #row index
for p in range(data_num): # column index
# Call `scatter` or any plot function on the
# respective `axes` object itself
axes[n,p].scatter(x[n,p], y[n,p])
print(f'plotting at index [ {n} , {p}]')
plt.savefig('figure.png')
and figure.png looks like (sorry, no colour or symbol variation, just bare bones scatter plots):
Brilliant, it was literally a case of changing that one line from 'axes[n,p] = plt.xcorr .......' to 'axes[n,p].xcorr(....' Really helpful, thank you!
– OsDavy
Nov 20 at 13:08
I corrected the part about the explanation.
– ImportanceOfBeingErnest
Nov 20 at 13:56
add a comment |
up vote
0
down vote
accepted
It's a, perhaps unfortunate, way that pyplot and matplotlib works: you have to create the plots on the respective axes, not assign the result from a pyplot.xcorr
call to an axes. Thus: axis[n,p].xcorr(...)
. So the interface is suddenly somewhat more object-oriented than the usual direct pyplot
calls.
All the plots ends up in just the last figure, because you are calling
plt.xcorr(x,y,normed=True)
It doesn't matter if you then assign the return value to the axes array elements, which you shouldn't, as that destroys the original axes array.plt.xcorr
will then plot all the data in the same plot on top of each other, because pyplot generally acts on the currently active axes, which is the last one created via plt.subplots()
.
That's for an explanation. Here's an example solution (with random data and a simple scatter plot):
import numpy as np
import matplotlib.pyplot as plt
data_num = 3
x = np.random.uniform(1, 10, size=(data_num, data_num, 20))
y = np.random.uniform(5, 20, size=(data_num, data_num, 20))
fig, axes = plt.subplots(nrows=data_num, ncols=data_num, sharex=True, sharey=True)
for n in range(data_num): #row index
for p in range(data_num): # column index
# Call `scatter` or any plot function on the
# respective `axes` object itself
axes[n,p].scatter(x[n,p], y[n,p])
print(f'plotting at index [ {n} , {p}]')
plt.savefig('figure.png')
and figure.png looks like (sorry, no colour or symbol variation, just bare bones scatter plots):
Brilliant, it was literally a case of changing that one line from 'axes[n,p] = plt.xcorr .......' to 'axes[n,p].xcorr(....' Really helpful, thank you!
– OsDavy
Nov 20 at 13:08
I corrected the part about the explanation.
– ImportanceOfBeingErnest
Nov 20 at 13:56
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
It's a, perhaps unfortunate, way that pyplot and matplotlib works: you have to create the plots on the respective axes, not assign the result from a pyplot.xcorr
call to an axes. Thus: axis[n,p].xcorr(...)
. So the interface is suddenly somewhat more object-oriented than the usual direct pyplot
calls.
All the plots ends up in just the last figure, because you are calling
plt.xcorr(x,y,normed=True)
It doesn't matter if you then assign the return value to the axes array elements, which you shouldn't, as that destroys the original axes array.plt.xcorr
will then plot all the data in the same plot on top of each other, because pyplot generally acts on the currently active axes, which is the last one created via plt.subplots()
.
That's for an explanation. Here's an example solution (with random data and a simple scatter plot):
import numpy as np
import matplotlib.pyplot as plt
data_num = 3
x = np.random.uniform(1, 10, size=(data_num, data_num, 20))
y = np.random.uniform(5, 20, size=(data_num, data_num, 20))
fig, axes = plt.subplots(nrows=data_num, ncols=data_num, sharex=True, sharey=True)
for n in range(data_num): #row index
for p in range(data_num): # column index
# Call `scatter` or any plot function on the
# respective `axes` object itself
axes[n,p].scatter(x[n,p], y[n,p])
print(f'plotting at index [ {n} , {p}]')
plt.savefig('figure.png')
and figure.png looks like (sorry, no colour or symbol variation, just bare bones scatter plots):
It's a, perhaps unfortunate, way that pyplot and matplotlib works: you have to create the plots on the respective axes, not assign the result from a pyplot.xcorr
call to an axes. Thus: axis[n,p].xcorr(...)
. So the interface is suddenly somewhat more object-oriented than the usual direct pyplot
calls.
All the plots ends up in just the last figure, because you are calling
plt.xcorr(x,y,normed=True)
It doesn't matter if you then assign the return value to the axes array elements, which you shouldn't, as that destroys the original axes array.plt.xcorr
will then plot all the data in the same plot on top of each other, because pyplot generally acts on the currently active axes, which is the last one created via plt.subplots()
.
That's for an explanation. Here's an example solution (with random data and a simple scatter plot):
import numpy as np
import matplotlib.pyplot as plt
data_num = 3
x = np.random.uniform(1, 10, size=(data_num, data_num, 20))
y = np.random.uniform(5, 20, size=(data_num, data_num, 20))
fig, axes = plt.subplots(nrows=data_num, ncols=data_num, sharex=True, sharey=True)
for n in range(data_num): #row index
for p in range(data_num): # column index
# Call `scatter` or any plot function on the
# respective `axes` object itself
axes[n,p].scatter(x[n,p], y[n,p])
print(f'plotting at index [ {n} , {p}]')
plt.savefig('figure.png')
and figure.png looks like (sorry, no colour or symbol variation, just bare bones scatter plots):
edited Nov 20 at 13:56
ImportanceOfBeingErnest
123k10127203
123k10127203
answered Nov 20 at 13:03
9769953
1,344311
1,344311
Brilliant, it was literally a case of changing that one line from 'axes[n,p] = plt.xcorr .......' to 'axes[n,p].xcorr(....' Really helpful, thank you!
– OsDavy
Nov 20 at 13:08
I corrected the part about the explanation.
– ImportanceOfBeingErnest
Nov 20 at 13:56
add a comment |
Brilliant, it was literally a case of changing that one line from 'axes[n,p] = plt.xcorr .......' to 'axes[n,p].xcorr(....' Really helpful, thank you!
– OsDavy
Nov 20 at 13:08
I corrected the part about the explanation.
– ImportanceOfBeingErnest
Nov 20 at 13:56
Brilliant, it was literally a case of changing that one line from 'axes[n,p] = plt.xcorr .......' to 'axes[n,p].xcorr(....' Really helpful, thank you!
– OsDavy
Nov 20 at 13:08
Brilliant, it was literally a case of changing that one line from 'axes[n,p] = plt.xcorr .......' to 'axes[n,p].xcorr(....' Really helpful, thank you!
– OsDavy
Nov 20 at 13:08
I corrected the part about the explanation.
– ImportanceOfBeingErnest
Nov 20 at 13:56
I corrected the part about the explanation.
– ImportanceOfBeingErnest
Nov 20 at 13:56
add a comment |
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