Comparing two lists and making new list
Question:
So lets say I have two lists a=[1,2,3,4,5,6]
and b=[2,34,5,67,5,6]
I want to create a third list which will have 1
where elements are different in a
and b
and 0
when they are same, so above would be like c=[1,1,1,1,0,0]
Answers:
Try a list comprehension over elements of each pair of items in the list with zip
:
[ 0 if i == j else 1 for i,j in zip(a,b) ]
You can zip the lists and compare them in a list comprehension. This takes advantage of the fact that booleans are equivalent to 1 and 0 in python:
a=[1,2,3,4,5,6]
b=[2,34,5,67,5,6]
[int(m!=n) for m, n, in zip(a, b)]
# [1, 1, 1, 1, 0, 0]
Iterating with a for loop is an option, though list comprehension may be more efficient.
a=[1,2,3,4,5,6]
b=[2,34,5,67,5,6]
c=[]
for i in range(len(a)):
if a[i] == b[i]:
c.append(0)
else:
c.append(1)
print(c)
prints
[1, 1, 1, 1, 0, 0]
If you will have multiple vector operations and they should be fast. Checkout numpy
.
import numpy as np
a=[1,2,3,4,5,6]
b=[2,34,5,67,5,6]
a = np.array(a)
b = np.array(b)
c = (a != b).astype(int)
# array([1, 1, 1, 1, 0, 0])
idk if this is exactly what youre loocking for but this should work:
edidt: just found out that Joe Thor commented almost the exact same a few minutes earlier than me lmao
a = [1, 2, 3, 4, 5, 6]
b = [2, 34, 5, 67, 5, 6]
results = []
for f in range(0, len(a)):
if a[f] == b[f]:
results.append(0)
else:
results.append(1)
print(results)
This can be done fairly simply using a for loop. It does assume that both lists, a and b, are the same length. An example code would like something like this:
a = [1,2,3,4,5,6]
b = [2,34,5,67,5,6]
c = []
if len(a) == len(b):
for i in range(0,len(a)):
if(a[i] != b[i]):
c.append(1)
else:
c.append(0)
This can also be done using list comprehension:
a = [1,2,3,4,5,6]
b = [2,34,5,67,5,6]
c = []
if len(a) == len(b):
c = [int(i != j) for i,j in zip(a,b)]
The list comprehension code is from this thread: Comparing values in two lists in Python
a = [1, 2, 3, 4, 5, 6]
b = [2, 34, 5, 67, 5,6]
c = []
index = 0
x = 1
y = 0
for i in range(len(a)): # iterating loop from index 0 till the last
if a[index]!= b[index]: # comapring each index
c.append(x) # if not equal append c with '1'
index += 1 # increment index to move to next index in both lists
else:
c.append(y)
index += 1
print(c)
This should work for two lists of any type.
tstlist = ["w","s","u"]
lstseasons = ["s","u","a","w"]
lstbool_Seasons = [1 if ele in tstlist else 0 for ele in lstseasons]
Output: lstbool_Seasons = [1,1,0,1]
This is the first time I have posted anything, still figuring out how things work here, so please forgive faux pas…
So lets say I have two lists a=[1,2,3,4,5,6]
and b=[2,34,5,67,5,6]
I want to create a third list which will have 1
where elements are different in a
and b
and 0
when they are same, so above would be like c=[1,1,1,1,0,0]
Try a list comprehension over elements of each pair of items in the list with zip
:
[ 0 if i == j else 1 for i,j in zip(a,b) ]
You can zip the lists and compare them in a list comprehension. This takes advantage of the fact that booleans are equivalent to 1 and 0 in python:
a=[1,2,3,4,5,6]
b=[2,34,5,67,5,6]
[int(m!=n) for m, n, in zip(a, b)]
# [1, 1, 1, 1, 0, 0]
Iterating with a for loop is an option, though list comprehension may be more efficient.
a=[1,2,3,4,5,6]
b=[2,34,5,67,5,6]
c=[]
for i in range(len(a)):
if a[i] == b[i]:
c.append(0)
else:
c.append(1)
print(c)
prints
[1, 1, 1, 1, 0, 0]
If you will have multiple vector operations and they should be fast. Checkout numpy
.
import numpy as np
a=[1,2,3,4,5,6]
b=[2,34,5,67,5,6]
a = np.array(a)
b = np.array(b)
c = (a != b).astype(int)
# array([1, 1, 1, 1, 0, 0])
idk if this is exactly what youre loocking for but this should work:
edidt: just found out that Joe Thor commented almost the exact same a few minutes earlier than me lmao
a = [1, 2, 3, 4, 5, 6]
b = [2, 34, 5, 67, 5, 6]
results = []
for f in range(0, len(a)):
if a[f] == b[f]:
results.append(0)
else:
results.append(1)
print(results)
This can be done fairly simply using a for loop. It does assume that both lists, a and b, are the same length. An example code would like something like this:
a = [1,2,3,4,5,6]
b = [2,34,5,67,5,6]
c = []
if len(a) == len(b):
for i in range(0,len(a)):
if(a[i] != b[i]):
c.append(1)
else:
c.append(0)
This can also be done using list comprehension:
a = [1,2,3,4,5,6]
b = [2,34,5,67,5,6]
c = []
if len(a) == len(b):
c = [int(i != j) for i,j in zip(a,b)]
The list comprehension code is from this thread: Comparing values in two lists in Python
a = [1, 2, 3, 4, 5, 6]
b = [2, 34, 5, 67, 5,6]
c = []
index = 0
x = 1
y = 0
for i in range(len(a)): # iterating loop from index 0 till the last
if a[index]!= b[index]: # comapring each index
c.append(x) # if not equal append c with '1'
index += 1 # increment index to move to next index in both lists
else:
c.append(y)
index += 1
print(c)
This should work for two lists of any type.
tstlist = ["w","s","u"]
lstseasons = ["s","u","a","w"]
lstbool_Seasons = [1 if ele in tstlist else 0 for ele in lstseasons]
Output: lstbool_Seasons = [1,1,0,1]
This is the first time I have posted anything, still figuring out how things work here, so please forgive faux pas…