here's my code:
d = ['ant', 'bird', 'gecko', 'snake', 'wasp']
a = ['rabbit panda bird rabbit', 'bird gecko ant panda', 'wasp snake gecko ant']
b = []
for i in range (0, len(a)):
c = a[i].split()
for i in d:
b.append(c.count(i))
print(b)
here's the output:
[0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1]
but i want the desired output like this:
[[0, 1, 0, 0, 0], [1, 1, 1, 0, 0], [1, 0, 1, 1, 1]]
I think this achieves what you want:
d = ['ant', 'bird', 'gecko', 'snake', 'wasp']
a = ['rabbit panda bird rabbit', 'bird gecko ant panda', 'wasp snake gecko ant']
b = []
for i in range (0, len(a)):
c = a[i].split()
e = []
for i in d:
e.append(c.count(i))
b.append(e)
print(b)
Note the addition of a local array 'e' which has the contents appended to, this is then appended to your array 'b'
Use List Comprehension:
d = ['ant', 'bird', 'gecko', 'snake', 'wasp']
a = ['rabbit panda bird rabbit', 'bird gecko ant panda', 'wasp snake gecko ant']
list_of_lists = [[x.split().count(y) for y in d] for x in a]
print list_of_lists
Output:
[[0, 1, 0, 0, 0], [1, 1, 1, 0, 0], [1, 0, 1, 1, 1]]
Using python's list comprehension
d = ['ant', 'bird', 'gecko', 'snake', 'wasp']
a = ['rabbit panda bird rabbit', 'bird gecko ant panda', 'wasp snake gecko ant']
b = []
for i in range (0, len(a)):
c = a[i].split()
b.append([c.count(i) for i in d])
print(b) # [[0, 1, 0, 0, 0], [1, 1, 1, 0, 0], [1, 0, 1, 1, 1]]
spam = ['ant', 'bird', 'gecko', 'snake', 'wasp']
eggs = ['rabbit panda bird rabbit', 'bird gecko ant panda', 'wasp snake gecko ant']
print([[int(word in egg.split(' ')) for word in spam] for egg in eggs])
output
[[0, 1, 0, 0, 0], [1, 1, 1, 0, 0], [1, 0, 1, 1, 1]]
Here you go -
d = ['ant', 'bird', 'gecko', 'snake', 'wasp']
a = ['rabbit panda bird rabbit', 'bird gecko ant panda', 'wasp snake gecko ant']
b = []
for i in range (0, len(a)):
c = a[i].split()
count_list = [c.count(i) for i in d]
b.append(count_list)
print(b)
Output -
[[0, 1, 0, 0, 0], [1, 1, 1, 0, 0], [1, 0, 1, 1, 1]]
The problem with your code is that you are not creating nor adding elements to a nested list. Here's a way to do it adapting your code:
# Start by creating a nsted list the same length of a
b = [[] for _ in range(len(a))]
# [[], [], []]
# Do the same but instead appending the elements to the sublists
# in b using i as index
for i in range (0, len(a)):
c = a[i].split()
for j in d:
b[i].append(c.count(j))
print(b)
# [[0, 1, 0, 0, 0], [1, 1, 1, 0, 0], [1, 0, 1, 1, 1]]
You could also use a nested list comprehension to achieve this:
[[j.split().count(i) for i in d] for j in a]
Output
[[0, 1, 0, 0, 0], [1, 1, 1, 0, 0], [1, 0, 1, 1, 1]]
Other one liner option:
[ [ sub.count(word) for word in d ] for sub in [ string.split() for string in a ] ]
#=> [[0, 1, 0, 0, 0], [1, 1, 1, 0, 0], [1, 0, 1, 1, 1]]
It states for each word in d count its occurrence in sub, where sub is a sub-list derived for each string in a.
[ string.split() for string in a ]
#=> [['rabbit', 'panda', 'bird', 'rabbit'], ['bird', 'gecko', 'ant', 'panda'], ['wasp', 'snake', 'gecko', 'ant']]
