I have two lists I1 and I2. I am comparing elements of I2 to elements in I1 and want to add the number of elements which are less than or equal to it. For example, 11 in I2 has two two elements less than or equal to it in I1. Hence, 2 should be added to 11. Similarly, for elements 27 and 41 in I2. I present the current and expected output.
I1=[1,11,13,16,26,30,43,46]
I2=[11,27,41]
I2=[i for i in I2 if I2<=I1]
print("I2 =",I2)
The current output is
I2 = []
The expected output is
I2 = [13,32,47]
CodePudding user response：
Try this.
lst1 = [1,11,13,16,26,30,43,46]
lst2 = [11,27,41]
lst2 = [a sum(a>=b for b in lst1) for a in lst2]
print("I2 =", lst2)
Output
I2 = [13, 32, 47]
Note: "Python variables should begin with lowercase letters (preferably ones that aren't easily mistaken for numbers like 1)" as @Chris says.
CodePudding user response：
use a counter to see the number and their frequency which is less than current, so to avoid the repetiiton calculation
from collections import Counter
I1=[1,11,13,16,26,30,43,46]
I2=[11,27,41]
c1 = Counter(I1)
result = []
for i in I2:
val = sum(j for v, j in c1.items() if v<=i) i
result.append(val)
print(result) # 13 32 47
CodePudding user response：
I am surprised thta nobody mentioned numpy masking yet.
data1 <= itemwill give you an array of booleans showing where that condition is true/false.(data1 <= item).sum()takes advantage of how booleans sum to integers in python to count the number of true items. You could uselen(data1[(data1 <= item)])but it's longer.- We do that for each item in your second array.
import numpy as np
data1 = np.array([1, 11, 13, 16, 26, 30, 43, 46])
data2 = np.array([11, 27, 41])
smaller_counts = np.array([(data1 <= item).sum() for item in data2])
# >>> [2 5 6]
result = data2 smaller_counts
# >>> [13 32 47]
CodePudding user response：
If I1 is always sorted and could be very large, you could use binary search for better performance, with the bisect module:
>>> import bisect
>>> [x bisect.bisect_right(I1, x) for x in I2]
[13, 32, 47]