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I have 3 lists with 3 sets of integers. I want to predict the outcome numbers using a list or by whatever other means is out there.
List1 = [0 0 2 0 1 2]
List2 = [3 0 1 0 1 0]
List3 = [0 0 2 1 1 1]
Im working in the first column, I was thinking of using a double if statement something like:
if 3 is in spot 0 of any list
and 0 is in spot 0 of any list
print 2
The other thing is I want to do this for all the other columns as well.
I know this is not complete but its on my mind like this and ive been
searching around and found nothing to solve this problem.
I would greatly appeciate any response to this question.
You cant define a list delimited with space; you have to use a comma',' instead.
l1=[0,0,2,0,1,2]
l2=[3,0,1,0,1,0,]
l3=[0,0,2,1,1,1]
spot0_list = zip(l1,l2,l3)[0]
if 3 in spot0_list and 0 in spot0_list:
print 2
And, if there are too many lists, you can use set:
spot0_list = set(zip(l1,l2,l3)[0])
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i need to convert this over to a recursive function. the program basically prints out all variations of a string.
def comb(L):
for i in range(3):
for j in range(3):
for k in range(3):
# check if the indexes are not
# same
if (i!=j and j!=k and i!=k):
print(L[i], L[j], L[k])
# Driver Code
comb([1, 2, 3])
output:
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
This might get you started:
Every permutation of a list can be made by picking a value from the list & putting it at the front of every permutation of what is left in the list after taking that value out. (Note that the "what is left" part is a smaller list than what you started with.)
When your list is small enough, it is the only permutation.
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I am trying to determine how many sentences there are in each row.
Sent
I went out for a walk.
I don't know. I think you're right!
so boring!!!
WTF?
Nothing
I created a list of punctuation symbols I am interested in for determining the number of sentences per each row:
Output
1
2
1
1
1
In order to get this result, I first considered to split each row whether I met a symbol (for instance . or ! or ?). But I do not know how to get the count.
My code is
import re
def sentence(sent):
return re.findall('[\w][\.!\?]', sent)
df['Sent'] = df['Sent'].apply(sentence)
Could you please give my advice on how to get it?
One idea if dont need last value like 1 use Series.str.count with regex for match letter with escaped .!?:
df['Output'] = df['Sent'].str.count('[\w][\.!\?]')
print (df)
Sent Output
0 I went out for a walk. 1
1 I don't know. I think you're right! 2
2 so boring!!! 1
3 WTF? 1
4 Nothing 0
If need replace 0 by 1:
df['Output'] = df['Sent'].str.count('[\w][\.!\?]').clip(lower=1)
print (df)
Sent Output
0 I went out for a walk. 1
1 I don't know. I think you're right! 2
2 so boring!!! 1
3 WTF? 1
4 Nothing 1
Another idea is use textstat lib:
import textstat
df['Output'] = df['Sent'].apply(textstat.sentence_count)
print (df)
Sent Output
0 I went out for a walk. 1
1 I don't know. I think you're right! 2
2 so boring!!! 1
3 WTF? 1
4 Nothing 1
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def multpoly(d1,d2):
d3=[]
d1=dict(d1)
d2=dict(d2)
for key,value in list(d1.items()):
for key2,value2 in list(d2.items()):
#print(key,value)
print(key2,value2)
print(key,value)
d1={1:1,-1:0}
d2={1:2,1:1,1:0}
wrong output:
1 0
1 1
1 0
-1 0
Expected:
1 2
1,1
1,0
1,1
-1,0
I am getting wrong output with this code?
Can anyone help me what is wrong with this code?
This is wrong with your code: dicts can only contain each key once.
If you specify
d1={1:1,-1:0}
d2={1:2,1:1,1:0}
You get dict d1 with contains keys 1 and -1 with its values and you get dict d2 wich contains only 1 as key (with value of 0 as this one was specified last).
Test:
d1={1:1,-1:0}
d2={1:2,1:1,1:0}
print(d1) # {1: 1, -1: 0}
print(d2) # {1: 0} - both of 1:2,1:1 were overwritten by the last key 1:0 spec
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Let's say I have four lists:
l1=[3,5,6]
l2=[0,2]
l3=[3,4,3,1,2]
l4=[2,3,2]
And I want to print them like this:
2
1
6 3 2
5 2 4 3
3 0 3 2
Can anyone help me, please?
If you are using Python 2.x you can use izip_longest() from itertools:
for i in reversed([' '.join(map(str,i)) for i in izip_longest(l1, l2, l3, l4, fillvalue=' ')]):
print i
In Python 3.x you can use zip_longest() from itertools.
Output:
2
1
6 3 2
5 2 4 3
3 0 3 2
you may be able to solve this problem by using multi-dimensional arrays to store the lists. then cycling through them checking for values to display while using an if statement to print a space if there is no value currently residing there (aka initialize the array to none where no value resides). I believe that if you structure it like that you may have some success
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I am looking for an efficient way to compare rows and columns of a table against each other (>= gets 1, otherwise 0) and store the result.
Example:
0.3642286 0.7945753 0.3527125
0.3642286 1 1 0
0.7945753 0 1 0
0.3527125 1 1 1
I have 21 tables with 480*480 rows and columns. What would be a proper way of generating and storing such a matrix?
All you really need is two loops.
def compare(first, second):
result = []
for x in first:
result.append([])
for y in second:
result[-1].append(1 if x >= y else 0)
result = [list(i) for i in zip(*result)]
return result
You might consider NumPy (1) if you are regularly handling large multi-dimensional arrays.