I have a list like below and need to firs add items in each list and then multiply all results 2+4 = 6 , 3+ (-2)=1, 2+3+2=7, -7+1=-6 then 6*1*7*(-6) = -252 I know how to do it by accessing indexes and it works (as below) but I also need to do it in a way that it will work no matter how many sublist there is
nested_lst = [[2,4], [3,-2],[2,3,2], [-7,1]]
a= nested_lst[0][0] + nested_lst[0][1]
b= nested_lst[1][0] + nested_lst[1][1]
c= nested_lst[2][0] + nested_lst[2][1] + nested_lst[2][2]
d= nested_lst[3][0] + nested_lst[3][1]
def sum_then_product(list):
multip= a*b*c*d
return multip
print sum_then_product(nested_lst)
I have tried with for loop which gives me addition but I don't know how to perform here multiplication. I am new to it. Please, help
nested_lst = [[2,4], [3,-2],[2,3,2], [-7,1]]
for i in nested_lst:
print sum(i)
Is this what you are looking for?
nested_lst = [[2,4], [3,-2],[2,3,2], [-7,1]] # your list
output = 1 # this will generate your eventual output
for sublist in nested_lst:
sublst_out = 0
for x in sublist:
sublst_out += x # your addition of the sublist elements
output *= sublst_out # multiply the sublist-addition with the other sublists
print(output)
Related
I want to count total number of '1's in binary format of a number which is in a list.
z = ['0b111000','0b1000011'] # z is a list
d = z.count('1')
print(d)
The output is 0.
Whereas the required output should be in the form of [3,3]
which is number of ones in every element that Z is containing :
Here it is :
z=['0b111000','0b1000011']
finalData = []
for word in z:
finalData.append(word.count('1'))
print(finalData)
The problem with your code was you were trying to use count() method on list type and it is used for string. You first need to get the string from the list and then use count() method on it.
Hope this helps :)
z = ['0b111000','0b1000011']
d = z.count('1')
This attempts to find the number of times the string '1' is in z. This obviously returns 0 since z contains '0b111000' and '0b1000011'.
You should iterate over every string in z and count the numbers of '1' in every string:
z = ['0b111000','0b1000011']
output = [string.count('1') for string in z]
print(output)
# [3, 3]
list.count(x) will count the number of occurrences such that it only counts the element if it is equal to x.
Use list comprehension to loop through each string and then count the number of 1s. Such as:
z = ['0b111000','0b1000011']
d = [x.count("1") for x in z]
print(d)
This will output:
[3, 3]
Assume I have the following matrix:
X = np.array([[1,2,3], [4,5,6], [7,8,9], [70,80,90], [45,43,68], [112,87,245]])
I want to draw a batch of 2 random rows at each time loop, and send it to a function. For instance, a batch in iteration i can be batch = [[4,5,6], [70,80,90]]
I do the following:
X = np.array([[1,2,3], [4,5,6], [7,8,9], [70,80,90], [45,43,68], [112,87,245]])
def caclulate_batch(batch):
pass
for i in range(X.shape[0]/2):
batch = np.array([])
for _ in range(2):
r = random.randint(0, 5)
batch = np.append(batch, X[r])
caclulate_batch(batch)
There are two problems here: (1) It returns appended array (2) The random number can be repeated which can choose the same row many times. How can modify the code to fit my requirement.
r = np.random.randint(0, len(x), 2) should get you the indices. That lets you use fancy indexing to get the subset: batch = x[r, :].
If you want to accumulate arrays along a new dimension, as your loop does, use np.stack or np.block instead of np.append.
(1) You can use numpy.stack instead of append. EDIT: But this function would be called when you have all your batch in a list like:
list = ([1,2], [3,4])
numpy.stack(list)
# gives [[1,2],
# [3,4]]
(2) You can shuffle X array, loop through the results and extract two by two. Look at numpy.random.shuffle
It would look like that:
S = np.random.shuffle(X)
for i in range(S.shape[0]/2):
batch = S[i*2:i*2+1]
caclulate_batch(batch)
I have this list: l = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20] and I would like to iterate on it, printing something like
0: [1,2,3,4]
1: [2,3,4,5]
2: [3,4,5,6]
...
n: [17,18,19,20]
So far I made this code to print 5 elements at a time, but the last iteration prints 3:
for index, item in enumerate(l):
if index == 0 or index == 1 or index == 2:
continue
print index, l[index - 3:index + 2]
How can I solve this?
You're on the right track with your list slices. Here's a tweak to make it easier:
sub_len = 4
for i in range(len(mylist) - sub_len + 1):
print l[i:i + sub_len]
Where sub_len is the desired length of the slices you print.
demo
trying to write this code to see how many times the numbers in list 1 appear in the list two, can use a nested for or while loop but I came up with this it doesn't work. I don't want to use count.
list1 = [4,7,2]
list2 = [2,3,4,2,5,6,3,2,6,7,3,4]
def compare(list1, list2):
freq = ([i for i in list1 if i == num])
return
print('The number 4 occurs in list2', freq, 'times')
print('The number 7 occurs in list2', freq, 'times')
print('The number 2 occurs in list2', freq, 'times')
I'm not completely sure that I understand the question,
but this code seems to work, though if it may be slow if you need it for an interactive program.
Hope this helps!
list1 = [4,7,2]
list2 = [2,3,4,2,5,6,3,2,6,7,3,4]
occurrences = [0,0,0]
for i in range(len(list1)):
for j in list2:
if list1[i] == j:
occurrences[i]+=1
print occurrences
try this:
list1 = [4,7,2]
list2 = [2,3,4,2,5,6,3,2,6,7,3,4]
occurrences = [0,0,0]
for i in range(len(list1)):
for j in list2:
if list1[i] == j:
occurrences[i]+=1
print occurrences
I'm working through the book NLP with Python, and I came across this example from an 'advanced' section. I'd appreciate help understanding how it works. The function computes all possibilities of a number of syllables to reach a 'meter' length n. Short syllables "S" take up one unit of length, while long syllables "L" take up two units of length. So, for a meter length of 4, the return statement looks like this:
['SSSS', 'SSL', 'SLS', 'LSS', 'LL']
The function:
def virahanka1(n):
if n == 0:
return [""]
elif n == 1:
return ["S"]
else:
s = ["S" + prosody for prosody in virahanka1(n-1)]
l = ["L" + prosody for prosody in virahanka1(n-2)]
return s + l
The part I don't understand is how the 'SSL', 'SLS', and 'LSS' matches are made, if s and l are separate lists. Also in the line "for prosody in virahanka1(n-1)," what is prosody? Is it what the function is returning each time? I'm trying to think through it step by step but I'm not getting anywhere. Thanks in advance for your help!
Adrian
Let's just build the function from scratch. That's a good way to understand it thoroughly.
Suppose then that we want a recursive function to enumerate every combination of Ls and Ss to make a given meter length n. Let's just consider some simple cases:
n = 0: Only way to do this is with an empty string.
n = 1: Only way to do this is with a single S.
n = 2: You can do it with a single L, or two Ss.
n = 3: LS, SL, SSS.
Now, think about how you might build the answer for n = 4 given the above data. Well, the answer would either involve adding an S to a meter length of 3, or adding an L to a meter length of 2. So, the answer in this case would be LL, LSS from n = 2 and SLS, SSL, SSSS from n = 3. You can check that this is all possible combinations. We can also see that n = 2 and n = 3 can be obtained from n = 0,1 and n=1,2 similarly, so we don't need to special-case them.
Generally, then, for n ≥ 2, you can derive the strings for length n by looking at strings of length n-1 and length n-2.
Then, the answer is obvious:
if n = 0, return just an empty string
if n = 1, return a single S
otherwise, return the result of adding an S to all strings of meter length n-1, combined with the result of adding an L to all strings of meter length n-2.
By the way, the function as written is a bit inefficient because it recalculates a lot of values. That would make it very slow if you asked for e.g. n = 30. You can make it faster very easily by using the new lru_cache from Python 3.3:
#lru_cache(maxsize=None)
def virahanka1(n):
...
This caches results for each n, making it much faster.
I tried to melt my brain. I added print statements to explain to me what was happening. I think the most confusing part about recursive calls is that it seems to go into the call forward but come out backwards, as you may see with the prints when you run the following code;
def virahanka1(n):
if n == 4:
print 'Lets Begin for ', n
else:
print 'recursive call for ', n, '\n'
if n == 0:
print 'n = 0 so adding "" to below'
return [""]
elif n == 1:
print 'n = 1 so returning S for below'
return ["S"]
else:
print 'next recursivly call ' + str(n) + '-1 for S'
s = ["S" + prosody for prosody in virahanka1(n-1)]
print '"S" + each string in s equals', s
if n == 4:
print '**Above is the result for s**'
print 'n =',n,'\n', 'next recursivly call ' + str(n) + '-2 for L'
l = ["L" + prosody for prosody in virahanka1(n-2)]
print '\t','what was returned + each string in l now equals', l
if n == 4:
print '**Above is the result for l**','\n','**Below is the end result of s + l**'
print 'returning s + l',s+l,'for below', '\n','='*70
return s + l
virahanka1(4)
Still confusing for me, but with this and Jocke's elegant explanation, I think I can understand what is going on.
How about you?
Below is what the code above produces;
Lets Begin for 4
next recursivly call 4-1 for S
recursive call for 3
next recursivly call 3-1 for S
recursive call for 2
next recursivly call 2-1 for S
recursive call for 1
n = 1 so returning S for below
"S" + each string in s equals ['SS']
n = 2
next recursivly call 2-2 for L
recursive call for 0
n = 0 so adding "" to below
what was returned + each string in l now equals ['L']
returning s + l ['SS', 'L'] for below
======================================================================
"S" + each string in s equals ['SSS', 'SL']
n = 3
next recursivly call 3-2 for L
recursive call for 1
n = 1 so returning S for below
what was returned + each string in l now equals ['LS']
returning s + l ['SSS', 'SL', 'LS'] for below
======================================================================
"S" + each string in s equals ['SSSS', 'SSL', 'SLS']
**Above is the result for s**
n = 4
next recursivly call 4-2 for L
recursive call for 2
next recursivly call 2-1 for S
recursive call for 1
n = 1 so returning S for below
"S" + each string in s equals ['SS']
n = 2
next recursivly call 2-2 for L
recursive call for 0
n = 0 so adding "" to below
what was returned + each string in l now equals ['L']
returning s + l ['SS', 'L'] for below
======================================================================
what was returned + each string in l now equals ['LSS', 'LL']
**Above is the result for l**
**Below is the end result of s + l**
returning s + l ['SSSS', 'SSL', 'SLS', 'LSS', 'LL'] for below
======================================================================
This function says that:
virakhanka1(n) is the same as [""] when n is zero, ["S"] when n is 1, and s + l otherwise.
Where s is the same as the result of "S" prepended to each elements in the resulting list of virahanka1(n - 1), and l the same as "L" prepended to the elements of virahanka1(n - 2).
So the computation would be:
When n is 0:
[""]
When n is 1:
["S"]
When n is 2:
s = ["S" + "S"]
l = ["L" + ""]
s + l = ["SS", "L"]
When n is 3:
s = ["S" + "SS", "S" + "L"]
l = ["L" + "S"]
s + l = ["SSS", "SL", "LS"]
When n is 4:
s = ["S" + "SSS", "S" + "SL", "S" + "LS"]
l = ["L" + "SS", "L" + "L"]
s + l = ['SSSS", "SSL", "SLS", "LSS", "LL"]
And there you have it, step by step.
You need to know the results of the other function calls in order to calculate the final value, which can be pretty messy to do manually as you can see. It is important though that you do not try to think recursively in your head. This would cause your mind to melt. I described the function in words, so that you can see that these kind of functions is are descriptions, and not a sequence of commands.
The prosody you see, that is a part of s and l definitions, are variables. They are used in a list-comprehension, which is a way of building lists. I've described earlier how this list is built.