I have some discrete data in an array, such that:
arr = np.array([[1,1,1],[2,2,2],[3,3,3],[2,2,2],[1,1,1]])
whose plot looks like:
I also have an index array, such that each unique value in arr is associated with a unique index value, like:
ind = np.array([[1,1,1],[2,2,2],[3,3,3],[4,4,4],[5,5,5]])
What is the most pythonic way of converting arr from discrete values to continuous values, so that the array would look like this when plotted?:
therefore, interpolating between the discrete points to make continuous data
I found a solution to this if anyone has a similar issue. It is maybe not the most elegant so modifications are welcome:
def ref_linear_interp(x, y):
arr = []
ux=np.unique(x) #unique x values
for u in ux:
idx = y[x==u]
try:
min = y[x==u-1][0]
max = y[x==u][0]
except:
min = y[x==u][0]
max = y[x==u][0]
try:
min = y[x==u][0]
max = y[x==u+1][0]
except:
min = y[x==u][0]
max = y[x==u][0]
if min==max:
sub = np.full((len(idx)), min)
arr.append(sub)
else:
sub = np.linspace(min, max, len(idx))
arr.append(sub)
return np.concatenate(arr, axis=None).ravel()
y = np.array([[1,1,1],[2,2,2],[3,3,3],[2,2,2],[1,1,1]])
x = np.array([[1,1,1],[2,2,2],[3,3,3],[4,4,4],[5,5,5]])
z = np.arange(1, 16, 1)
Here is an answer for the symmetric solution that I would expect when reading the question:
import matplotlib.pyplot as plt
import pandas as pd
import numpy as np
# create the data as described
numbers = [1,2,3,2,1]
nblock = 3
df = pd.DataFrame({
"x": np.arange(nblock*len(numbers)),
"y": np.repeat(numbers, nblock),
"label": np.repeat(np.arange(len(numbers)), nblock)
})
Expecting a constant block size of 3, we could use a rolling window:
df['y-smooth'] = df['y'].rolling(nblock, center=True).mean()
# fill NaNs
df['y-smooth'].bfill(inplace=True)
df['y-smooth'].ffill(inplace=True)
plt.plot(df['x'], df['y-smooth'], marker='*')
If the block size is allowed to vary, we could determine the block centers and interpolate piecewise.
centers = df[['x', 'y', 'label']].groupby('label').mean()
df['y-interp'] = np.interp(df['x'], centers['x'], centers['y'])
plt.plot(df['x'], df['y-interp'], marker='*')
Note: You may also try
centers = df[['x', 'y', 'label']].groupby('label').min() to select the left corner of the labelled blocks.
Related
I have two numpy arrays:
left = np.array([2, 7])
right = np.array([4, 7])
right_p1 = right + 1
What I want to do is
rand = np.zeros(left.shape[0])
for i in range(left.shape[0]):
rand[i] = np.random.randint(left[i], right_p1[i])
Is there a way I could do this without using a for loop?
You could try with:
extremes = zip(left, right_p1)
rand = map(lambda x: np.random.randint(x[0], x[1]), extremes)
This way you will end up with a map object. If you need to save memory, you can keep it that way, otherwise you can get the full np.array passing through a list conversion, like this:
rand = np.array(list(map(lambda x: np.random.randint(x[0], x[1]), extremes)))
Suppose that I have an array
a = np.array([[1,2.5,3,4],[1, 2.5, 3,3]])
I want to find the mode of each column without using stats.mode().
The only way I can think of is the following:
result = np.zeros(a.shape[1])
for i in range(len(result)):
curr_col = a[:,i]
result[i] = curr_col[np.argmax(np.unique(curr_col, return_counts = True))]
update:
There is some error in the above code and the correct one should be:
values, counts = np.unique(a[:,i], return_counts = True)
result[i] = values[np.argmax(counts)]
I have to use the loop because np.unique does not output compatible result for each column and there is no way to use np.bincount because the dtype is not int.
If you look at the numpy.unique documentation, this function returns the values and the associated counts (because you specified return_counts=True). A slight modification of your code is necessary to give the correct result. What you are trying todo is to find the value associated to the highest count:
import numpy as np
a = np.array([[1,5,3,4],[1,5,3,3],[1,5,3,3]])
result = np.zeros(a.shape[1])
for i in range(len(result)):
values, counts = np.unique(a[:,i], return_counts = True)
result[i] = values[np.argmax(counts)]
print(result)
Output:
% python3 script.py
[1. 5. 3. 4.]
Here is a code tha compares your solution with the scipy.stats.mode function:
import numpy as np
import scipy.stats as sps
import time
a = np.random.randint(1,100,(100,100))
t_start = time.time()
result = np.zeros(a.shape[1])
for i in range(len(result)):
values, counts = np.unique(a[:,i], return_counts = True)
result[i] = values[np.argmax(counts)]
print('Timer 1: ', (time.time()-t_start), 's')
t_start = time.time()
result_2 = sps.mode(a, axis=0).mode
print('Timer 2: ', (time.time()-t_start), 's')
print('Matrices are equal!' if np.allclose(result, result_2) else 'Matrices differ!')
Output:
% python3 script.py
Timer 1: 0.002721071243286133 s
Timer 2: 0.003339052200317383 s
Matrices are equal!
I tried several values for parameters and your code is actually faster than scipy.stats.mode function so it is probably close to optimal.
How to find the correlation-peak values and coordinates of a set of 2D cross-correlation functions?
Given an 3D ndarray that contains a set of 2D cross-correlation functions. What is the efficient way to find the maximum(peak) values and their coordinates(x and y indices)?
The code below do the work but I think it is inefficient.
import numpy as np
import numpy.matlib
ccorr = np.random.rand(7,5,5)
xind = ccorr.argmax(axis=-1)
mccorr = ccorr[np.matlib.repmat(np.arange(0,7)[:,np.newaxis],1,5),np.matlib.repmat(np.arange(0,5)[np.newaxis,:],7,1), xind]
yind = mccorr.argmax(axis=-1)
xind = xind[np.arange(0,7),yind]
values = mccorr[np.arange(0,7),yind]
print("cross-correlation functions (z,y,x)")
print(ccorr)
print("x and y indices of the maximum values")
print(xind,yind)
print("Maximum values")
print(values)
You'll want to flatten the dimensions you're searching over and then use unravel_index and take_along_axis to get the coordinates and values, respectively.
ccorr = np.random.rand(7,5,5)
cc_rav = ccorr.reshape(ccorr.shape[0], -1)
idx = np.argmax(cc_rav, axis = -1)
indices_2d = np.unravel_index(idx, ccorr.shape[1:])
vals = np.take_along_axis(ccorr, indices = indices_2d, axis = 0)
if you're using numpy version <1.15:
vals = cc_rav[np.arange(ccorr.shape[0]), idx]
or:
vals = ccorr[np.arange(ccorr.shape[0]),
indices_2d[0], indices_2d[1]]
I have a data frame as following,
df.head()
ID AS_FP AC_FP RP11_FP RP11_be AC_be AS_be Info
AE02 0.060233 0 0.682884 0.817115 0.591182 0.129252 SAP
AE03 0 0 0 0.889181 0.670113 0.766243 SAP
AE04 0 0 0.033256 0.726193 0.171861 0.103839 others
AE05 0 0 0.034988 0.451329 0.431836 0.219843 others
What I am aiming is to plot each column starting from AS_FP til RP11_beta as lmplot, each x axis is column ending with FP and y axis is its corresponding column ending with be.
And I wanted to save it as separate files so I strated iterating through the columns by skipping first column ID, like this,
for ind, column in enumerate(df.columns):
if column.split('_')[0] == column.split('_')[0]:
But I got lost how to continue, I need to plot
sns.lmplot(x, y, data=df, hue='Info',palette=colors, fit_reg=False,
size=10,scatter_kws={"s": 700},markers=["o", "v"])
and save each image as seperate file
Straightforward solution:
1) Toy data:
import pandas as pd
from collections import OrderedDict
import matplotlib.pyplot as plt
import seaborn as sns
dct = OrderedDict()
dct["ID"] = ["AE02", "AE03", "AE04", "AE05"]
dct["AS_FP"] = [0.060233, 0, 0, 0]
dct["AC_FP"] = [0, 0,0, 0]
dct["RP11_FP"] = [0.682884, 0, 0.033256, 0.034988]
dct["AS_be"] = [0.129252, 0.766243, 0.103839, 0.219843]
dct["AC_be"] = [0.591182, 0.670113, 0.171861, 0.431836]
dct["RP11_be"] = [0.817115, 0.889181, 0.726193, 0.451329]
dct["Info"] = ["SAP", "SAP", "others", "others"]
df = pd.DataFrame(dct)
2) Iterating through pairs, saving each figure with unique filename:
graph_cols = [col for col in df.columns if ("_FP" in col) or ("_be" in col)]
fps = sorted([col for col in graph_cols if "_FP" in col])
bes = sorted([col for col in graph_cols if "_be" in col])
for x, y in zip(fps, bes):
snsplot = sns.lmplot(x, y, data=df, fit_reg=False, hue='Info',
size=10, scatter_kws={"s": 700})
snsplot.savefig(x.split("_")[0] + ".png")
You can add needed params in lmlplot as you need.
I am trying to vectorize my code and, thanks in large part to some users (https://stackoverflow.com/users/3293881/divakar, https://stackoverflow.com/users/625914/behzad-nouri), I was able to make huge progress. Essentially, I am trying to apply a generic function (in this case max_dd_array_ret) to each of the bins I found (see vectorize complex slicing with pandas dataframe for details on date vectorization and Start, End and Duration of Maximum Drawdown in Python for the rationale behind max_dd_array_ret). the problem is the following: I should be able to obtain the result df_2 and, to some degree, ranged_DD(asd_1.values, starts, ends+1) is what I am looking for, except for the tragic effect that it's as if the first two bins are merged and the last one is missing as it can be gauged by looking at the results.
any explanation and fix is very welcomed
import pandas as pd
import numpy as np
from time import time
from scipy.stats import binned_statistic
def max_dd_array_ret(xs):
xs = (xs+1).cumprod()
i = np.argmax(np.maximum.accumulate(xs) - xs) # end of the period
j = np.argmax(xs[:i])
max_dd = abs(xs[j]/xs[i] -1)
return max_dd if max_dd is not None else 0
def get_ranges_arr(starts,ends):
# Taken from https://stackoverflow.com/a/37626057/3293881
counts = ends - starts
counts_csum = counts.cumsum()
id_arr = np.ones(counts_csum[-1],dtype=int)
id_arr[0] = starts[0]
id_arr[counts_csum[:-1]] = starts[1:] - ends[:-1] + 1
return id_arr.cumsum()
def ranged_DD(arr,starts,ends):
# Get all indices and the IDs corresponding to same groups
idx = get_ranges_arr(starts,ends)
id_arr = np.repeat(np.arange(starts.size),ends-starts)
slice_arr = arr[idx]
return binned_statistic(id_arr, slice_arr, statistic=max_dd_array_ret)[0]
asd_1 = pd.Series(0.01 * np.random.randn(500), index=pd.date_range('2011-1-1', periods=500)).pct_change()
index_1 = pd.to_datetime(['2011-2-2', '2011-4-3', '2011-5-1','2011-7-2', '2011-8-3', '2011-9-1','2011-10-2', '2011-11-3', '2011-12-1','2012-1-2', '2012-2-3', '2012-3-1',])
index_2 = pd.to_datetime(['2011-2-15', '2011-4-16', '2011-5-17','2011-7-17', '2011-8-17', '2011-9-17','2011-10-17', '2011-11-17', '2011-12-17','2012-1-17', '2012-2-17', '2012-3-17',])
starts = asd_1.index.searchsorted(index_1)
ends = asd_1.index.searchsorted(index_2)
df_2 = pd.DataFrame([max_dd_array_ret(asd_1.loc[i:j]) for i, j in zip(index_1, index_2)], index=index_1)
print(df_2[0].values)
print(ranged_DD(asd_1.values, starts, ends+1))
results:
df_2
[ 1.75893509 6.08002911 2.60131797 1.55631781 1.8770067 2.50709085
1.43863472 1.85322338 1.84767224 1.32605754 1.48688414 5.44786663]
ranged_DD(asd_1.values, starts, ends+1)
[ 6.08002911 2.60131797 1.55631781 1.8770067 2.50709085 1.43863472
1.85322338 1.84767224 1.32605754 1.48688414]
which are identical except for the first two:
[ 1.75893509 6.08002911 vs [ 6.08002911
and the last two
1.48688414 5.44786663] vs 1.48688414]
p.s.:while looking in more detail at the docs (http://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.binned_statistic.html) I found that this might be the problem
"All but the last (righthand-most) bin is half-open. In other words,
if bins is [1, 2, 3, 4], then the first bin is [1, 2) (including 1,
but excluding 2) and the second [2, 3). The last bin, however, is [3,
4], which includes 4. New in version 0.11.0."
problem is I don't how to reset it.