order_list_raw = []
for i in range(1, 73):
order_list_raw.append(1)
order_list_raw.append(2)
order_list_raw.append(3)
How can I create the same list with a randomized order but without having the same entry successively (e.g. "1, 3, 2" is okay but not "1, 1, 3").
For randomization I would create a new list like this:
order_list = random.sample(order_list_raw, len(order_list_raw))
A solution would be:
result = []
for i in range(72):
options = [1, 2, 3]
try:
last_item = result[-1]
options.remove(last_item)
except IndexError:
pass
result.append(random.choice(options))
print(result)
Output:
[1, 3, 2, 1, 2, 3, 1, 2, 3, 2, 1, 3, 2, 3, 2, 1, 2, 1, 2, 3, 1, 2, 1, 3, 1, 2, 3, 2, 3, 2, 3, 2, 1, 2, 3, 1, 2, 3, 2, 1, 2, 1, 3, 2, 3, 2, 3, 2, 1, 2, 3, 2, 3, 1, 3, 2, 1, 3, 1, 3, 1, 3, 1, 2, 3, 2, 1, 3, 1, 2, 1, 3]
Here we simply take our options, check what the last value in the list is and delete that value from the options. Then we take a random value from the left over options, and append it to the list.
In case if you want to generate the input data randomly then you can use this solution.
import random
b=[]
for i in range(0,73):
x=random.randint(1,10)
if len(b)==0 or b[-1]!=x:
b.append(x)
print(b)
Output :
[6, 2, 3, 5, 6, 5, 3, 8, 1, 5, 4, 9, 4, 9, 8, 6, 9, 2, 1, 5, 8, 6, 1, 9, 6, 9, 3, 6, 5, 7, 9, 1, 9, 5, 9, 3, 4, 3, 7, 8, 3, 4, 5, 9, 1, 4, 9, 2, 1, 5, 7, 1, 10, 2, 4, 2, 1, 7, 1, 5, 4, 1, 2]
But in case if your input data is fixed, then you can try this solution as below.
a=[1,1,4]
b=[]
c=[[b.append(i) for i in a if len(b)==0 or b[-1]!=i]for j in range(0,100)]
print(b)
Output :
[1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4]
Related
I want to generate random result with weights based on the previous result.
I am generating random results with weights in the following manner
a = [1, 2, 3, 4, 5, 6]
results = random.choices(a, weights=(
(10, 20, 30, 30, 10, 6)), k=100)
print (results[0:100]
What I want to do is if results[n] = 1, the next result cannot be 6(i.e. it can be between 1 and 5).
I am new to this page and python. Any help would be useful
Using Kelly Bundy's suggestion from the comments:
a = [1, 2, 3, 4, 5, 6]
choices = random.choices(a, weights=(10, 20, 30, 30, 10, 6), k=100)
# print(choices)
for i, (prev, curr) in enumerate(zip(choices[:-1], choices[1:])):
if prev == 1 and curr == 6:
# print(f'Rerolled {i + 1}')
choices[i + 1] = random.choices(a, weights=(10, 20, 30, 30, 10, 0), k=1)[0]
print(choices)
Example Output:
[5, 4, 4, 1, 3, 4, 5, 6, 3, 4, 1, 2, 4, 2, 4, 3, 1, 3, 2, 3, 1, 2, 5, 4, 4, 3, 6, 4, 3, 4, 1, 5, 1, 3, 1, 3, 2, 1, 3, 4, 2, 4, 4, 1, 5, 2, 4, 4, 4, 6, 3, 3, 3, 2, 1, 4, 2, 2, 5, 4, 4, 2, 2, 4, 1, 3, 4, 4, 5, 4, 4, 4, 4, 3, 1, 4, 3, 4, 4, 4, 4, 4, 4, 2, 3, 5, 3, 2, 4, 4, 1, 2, 5, 6, 3, 4, 4, 6, 4, 3]
You could change the corresponding weight for 6 to 0 for the next choice if the previous choice was a 1:
a = [1, 2, 3, 4, 5, 6]
choices = []
k = 100
i = 0
previous_choice_1 = False
while i < k:
if previous_choice_1:
choice = random.choices(a, weights=(10, 20, 30, 30, 10, 0), k=1)[0]
else:
choice = random.choices(a, weights=(10, 20, 30, 30, 10, 6), k=1)[0]
choices.append(choice)
previous_choice_1 = choice == 1
i += 1
print(choices)
Example Output:
[6, 3, 3, 6, 2, 3, 3, 1, 2, 3, 6, 1, 2, 2, 6, 4, 3, 3, 2, 2, 1, 2, 3, 4, 4, 3, 2, 5, 4, 4, 3, 3, 4, 3, 3, 3, 1, 4, 5, 1, 2, 4, 4, 2, 4, 3, 4, 3, 6, 4, 1, 5, 3, 4, 4, 2, 4, 3, 4, 3, 3, 4, 3, 2, 2, 2, 3, 4, 3, 2, 1, 4, 4, 3, 3, 1, 3, 4, 5, 4, 4, 3, 1, 2, 2, 6, 3, 2, 3, 3, 4, 2, 1, 2, 4, 3, 3, 3, 1, 4]
I have a np.array q with some values for example: [1,3,5,7] .
And a np.array z. with some values that I need to round and than they are used as index in the
Third array 'mapping'.
import numpy as np
q = [1,3,5,7]
z = [0,50.3,240.4,252.9,256]
mapping = np.zeros(256)
for i in range(len(q)):
print(i)
start, end = int(round(z[i])), int(round(z[i + 1]))
mapping[start:end] = int(round(q[i]))
print(mapping)
The output here is:
Here's my approach:
repeats = np.diff(list(np.round(z))+ [256]).astype(int)
# repeats = array([ 49, 191, 12, 3])
np.repeat(np.round(q), repeats)
Output:
array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 5, 5,
5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 7, 7, 7])
Note: this only has 255 elements and it's different from your expected output, because, tbh I don't really understand your logic.
I have created a list
a=[1,2,3,4,5]*100
I now need to create another list that will contain the first 8 prime number locations from within a.
I have tried these two lines of code and they didn't work
b=a[2:3:5:7:11:13:17:19]
a[2:3:5:7:11:13:17:19]=b
The output for list A is "[1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5]" so its the locations 2,3,5,7,11,13,17,19 out of that output
a=[1,2,3,4,5]*100
indices = [2,3,5,7,11,13,17,19]
b = []
for i in indices:
b.append(a[i])
print(b)
You have to access each element individually. b=a[2:3:5:7:11:13:17:19] is not valid syntatically in Python. Actually, this is not the way to access elements at particular indices.
Pythonic way to do the same thing (It will reduce code length) using List Comprehension:
indices = [2,3,5,7,11,13,17,19]
b = [a[i] for i in indices]
I would try it like this using list comprehension (beware the test_prime method is not optimized at all):
def test_prime(n):
if (n==1):
return False
elif (n==2):
return True;
else:
for x in range(2,n):
if(n % x==0):
return False
return True
a=[1,2,3,4,5]*100
b = [item for item in range(len(a)) if test_prime(a[item])]
b = b[0:8]
print b
which outputs (note Python counts from 0, so the first element of an array is 0 and not 1):
[1, 2, 4, 6, 7, 9, 11, 12]
How would I find the values from a certain column in an array? For example I have:
[1, 1, 2, 4, 1, 7, 1, 7, 6, 9]
[1, 2, 5, 3, 9, 1, 1, 1, 9, 1]
[7, 4, 5, 1, 8, 1, 2, 0, 0, 4]
[1, 4, 1, 1, 1, 1, 1, 1, 8, 5]
[9, 0, 0, 0, 0, 0, 1, 1, 9, 8]
[7, 4, 2, 1, 8, 2, 2, 2, 9, 7]
[7, 4, 2, 1, 7, 1, 1, 1, 0, 5]
[3, 4, 5, 3, 4, 5, 9, 1, 0, 9]
[0, 0, 5, 1, 1, 1, 9, 7, 7, 7]
If I wanted to list all of the values of column 5, how would I do this? I have figured out how to do this for the rows, but for the columns it is tricky, since they are all part of a separate list. I have not been able to find anything about this and I am very new to Python so I don't really know what I don't know.
It's simple. Just use l[i][4] to print 5th column value.
l = [
[1, 1, 2, 4, 1, 7, 1, 7, 6, 9],
[1, 2, 5, 3, 9, 1, 1, 1, 9, 1],
[7, 4, 5, 1, 8, 1, 2, 0, 0, 4],
[1, 4, 1, 1, 1, 1, 1, 1, 8, 5],
[9, 0, 0, 0, 0, 0, 1, 1, 9, 8],
[7, 4, 2, 1, 8, 2, 2, 2, 9, 7],
[7, 4, 2, 1, 7, 1, 1, 1, 0, 5],
[3, 4, 5, 3, 4, 5, 9, 1, 0, 9],
[0, 0, 5, 1, 1, 1, 9, 7, 7, 7]
]
for i in l:
print(i[4])
# or simply use
[i[4] for i in l] #as pointed out by #COLDSPEED
# the above code will create a list with values from 5th column
See it in action here
For a two dimensional array, you can use array[row][column].
I have multidimensional data in a pandas data frame with one variable indicating class. For example here is my attempt with a poor-maps heatmap scatter plot:
import pandas as pd
import random
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.cm import get_cmap
nrows=1000
df=pd.DataFrame([[random.random(), random.random()]+[random.randint(0, 1)] for _ in range(nrows)],
columns=list("ABC"))
bins=np.linspace(0, 1, 20)
df["Abin"]=[bins[i-1] for i in np.digitize(df.A, bins)]
df["Bbin"]=[bins[i-1] for i in np.digitize(df.B, bins)]
g=df.ix[:,["Abin", "Bbin"]+["C"]].groupby(["Abin", "Bbin"])
data=g.agg(["sum", "count"])
data.reset_index(inplace=True)
data["classratio"]=data[("C", "sum")]/data[("C","count")]
plt.scatter(data.Abin, data.Bbin, c=data.classratio, cmap=get_cmap("RdYlGn_r"), marker="s")
I'd like to plot class densities over binned features. Now I used np.digitize for binning and some complicating Python hand-made density calculation to plot a heatmap.
Surely, this can be done more compactly with Pandas (pivot?)? Do you know a neat way to bin the two features (for example 10 bins on the interval 0...1) and then plot a class density heatmap where color indicates the ratio of 1's to total rows within this 2D-bin?
Yep, it can be done in a very concise way using the build in cut function:
In [65]:
nrows=1000
df=pd.DataFrame([[random.random(), random.random()]+[random.randint(0, 1)] for _ in range(nrows)],
columns=list("ABC"))
In [66]:
#This does the trick.
pd.crosstab(np.array(pd.cut(df.A, 20)), np.array(pd.cut(df.B, 20))).values
Out[66]:
array([[2, 2, 2, 2, 7, 2, 3, 5, 1, 4, 2, 2, 1, 3, 2, 1, 7, 2, 4, 2],
[1, 2, 4, 2, 0, 3, 3, 3, 1, 1, 2, 1, 4, 3, 2, 1, 1, 2, 2, 1],
[0, 4, 1, 3, 1, 3, 2, 5, 2, 3, 1, 1, 1, 4, 2, 3, 6, 5, 2, 2],
[5, 2, 3, 2, 2, 1, 3, 2, 4, 0, 3, 2, 0, 4, 3, 2, 1, 3, 1, 3],
[2, 2, 4, 1, 3, 2, 2, 4, 1, 4, 3, 5, 5, 2, 3, 3, 0, 2, 4, 0],
[2, 3, 3, 5, 2, 0, 5, 3, 2, 3, 1, 2, 5, 4, 4, 3, 4, 3, 6, 4],
[3, 2, 2, 4, 3, 3, 2, 0, 0, 4, 3, 2, 2, 5, 4, 0, 1, 2, 2, 3],
[0, 0, 4, 4, 3, 2, 4, 6, 4, 2, 0, 5, 2, 2, 1, 3, 4, 4, 3, 2],
[3, 2, 2, 3, 4, 2, 1, 3, 1, 3, 4, 2, 4, 3, 2, 3, 2, 3, 4, 4],
[0, 1, 1, 4, 1, 4, 3, 0, 1, 1, 1, 2, 6, 4, 3, 5, 3, 3, 1, 4],
[2, 2, 4, 1, 3, 4, 1, 2, 1, 3, 3, 3, 1, 2, 1, 5, 2, 1, 4, 3],
[0, 0, 0, 4, 2, 0, 2, 3, 2, 2, 2, 4, 4, 2, 3, 2, 1, 2, 1, 0],
[3, 3, 0, 3, 1, 5, 1, 1, 2, 5, 6, 5, 0, 0, 3, 2, 1, 5, 7, 2],
[3, 3, 2, 1, 2, 2, 2, 2, 4, 0, 1, 3, 3, 1, 5, 6, 1, 3, 2, 2],
[3, 0, 3, 4, 3, 2, 1, 4, 2, 3, 4, 0, 5, 3, 2, 2, 4, 3, 0, 2],
[0, 3, 2, 2, 1, 5, 1, 4, 3, 1, 2, 2, 3, 5, 1, 2, 2, 2, 1, 2],
[1, 3, 2, 1, 1, 4, 4, 3, 2, 2, 5, 5, 1, 0, 1, 0, 4, 3, 3, 2],
[2, 2, 2, 1, 1, 3, 1, 6, 5, 2, 5, 2, 3, 4, 2, 2, 1, 1, 4, 0],
[3, 3, 4, 7, 0, 2, 6, 4, 1, 3, 4, 4, 1, 4, 1, 1, 2, 1, 3, 2],
[3, 6, 3, 4, 1, 3, 1, 3, 3, 1, 6, 2, 2, 2, 1, 1, 4, 4, 0, 4]])
In [67]:
abins=np.linspace(df.A.min(), df.A.max(), 21)
bbins=np.linspace(df.B.min(), df.B.max(), 21)
Z=pd.crosstab(np.array(pd.cut(df.ix[df.C==1, 'A'], abins)),
np.array(pd.cut(df.ix[df.C==1, 'B'], bbins)), aggfunc=np.mean).div(
pd.crosstab(np.array(pd.cut(df.A, abins)),
np.array(pd.cut(df.B, bbins)), aggfunc=np.mean)).values
Z = np.ma.masked_where(np.isinf(Z),Z)
x=np.linspace(df.A.min(), df.A.max(), 20)
y=np.linspace(df.B.min(), df.B.max(), 20)
X,Y=np.meshgrid(x, y)
plt.contourf(X, Y, Z, vmin=0, vmax=1)
plt.colorbar()
plt.pcolormesh(X, Y, Z, vmin=0, vmax=1)
plt.colorbar()