I want to ask how to check if a variable is table 1x8 or 8x1 of type logical?
I know I can check the class of an array for logical like this:
strcmp(class(a),'logical')
I know I can get the size of table like this:
[h w] = size(a);
if(w==1 & h==8 | w==8 & h==1)
But what if table has more than 2 dimensions? How can I get number of dimensions?
To get the number of dimensions, use ndims
numDimensions = ndims(a);
However, you can instead request size to return a single output, which is an array [sizeX,sizeY,sizeZ,...], and check its length. Even better, you can use isvector to test whether it's a 1-d array.
So you can write
if isvector(a) && length(a) == 8
disp('it''s a 1x8 or 8x1 array')
end
Finally, to test for logical, it's easier to write
islogical(a)
Related
I have a dataframe which has a column called hexa which has hex values like this. They are of dtype object.
hexa
0 00802259AA8D6204
1 00802259AA7F4504
2 00802259AA8D5A04
I would like to remove the first and last bits and reverse the values bitwise as follows:
hexa-rev
0 628DAA592280
1 457FAA592280
2 5A8DAA592280
Please help
I'll show you the complete solution up here and then explain its parts below:
def reverse_bits(bits):
trimmed_bits = bits[2:-2]
list_of_bits = [i+j for i, j in zip(trimmed_bits[::2], trimmed_bits[1::2])]
reversed_bits = [list_of_bits[-i] for i in range(1,len(list_of_bits)+1)]
return ''.join(reversed_bits)
df['hexa-rev'] = df['hexa'].apply(lambda x: reverse_bits(x))
There are possibly a couple ways of doing it, but this way should solve your problem. The general strategy will be defining a function and then using the apply() method to apply it to all values in the column. It should look something like this:
df['hexa-rev'] = df['hexa'].apply(lambda x: reverse_bits(x))
Now we need to define the function we're going to apply to it. Breaking it down into its parts, we strip the first and last bit by indexing. Because of how negative indexes work, this will eliminate the first and last bit, regardless of the size. Your result is a list of characters that we will join together after processing.
def reverse_bits(bits):
trimmed_bits = bits[2:-2]
The second line iterates through the list of characters, matches the first and second character of each bit together, and then concatenates them into a single string representing the bit.
def reverse_bits(bits):
trimmed_bits = bits[2:-2]
list_of_bits = [i+j for i, j in zip(trimmed_bits[::2], trimmed_bits[1::2])]
The second to last line returns the list you just made in reverse order. Lastly, the function returns a single string of bits.
def reverse_bits(bits):
trimmed_bits = bits[2:-2]
list_of_bits = [i+j for i, j in zip(trimmed_bits[::2], trimmed_bits[1::2])]
reversed_bits = [list_of_bits[-i] for i in range(1,len(list_of_bits)+1)]
return ''.join(reversed_bits)
I explained it in reverse order, but you want to define this function that you want applied to your column, and then use the apply() function to make it happen.
My program --> I Will ask the user to introduce a number and I want to make that if the number is not in a random sequence (I choose 1,2,3) of numbers, the user need to write again a number until the number they enter is in the sequence:
a = (1,2,3)
option = int(input(''))
while option != a:
print('Enter a number between 1 and 3 !!')
option = int(input(''))
So as you can see I use the variable as a tuple but I don't know how to do it.. =(
Assuming the use of a tuple is obligatory, you will need to get input as a string, because it is iterable type. It will alow you easily convert to int, sign by sign, thru list comprehension. Now you have a list of ints, which you simply convert to a tuple. The final option variable looks:
option = tuple([int(sign) for sign in str(input(''))])
But consider keeping your signature in int instead of tuple. Int number is also unequivocal if its about sequence. In python 123 == 132 returns False. That way, you need only to replace:
a = (1,2,3)
by a:
a = 123
And script will works.
I'm very confused on the apply function for pandas. I have a big dataframe where one column is a column of strings. I'm then using a function to count part-of-speech occurrences. I'm just not sure the way of setting up my apply statement or my function.
def noun_count(row):
x = tagger(df['string'][row].split())
# array flattening and filtering out all but nouns, then summing them
return num
So basically I have a function similar to the above where I use a POS tagger on a column that outputs a single number (number of nouns). I may possibly rewrite it to output multiple numbers for different parts of speech, but I can't wrap my head around apply.
I'm pretty sure I don't really have either part arranged correctly. For instance, I can run noun_count[row] and get the correct value for any index but I can't figure out how to make it work with apply how I have it set up. Basically I don't know how to pass the row value to the function within the apply statement.
df['num_nouns'] = df.apply(noun_count(??),1)
Sorry this question is all over the place. So what can I do to get a simple result like
string num_nouns
0 'cat' 1
1 'two cats' 1
EDIT:
So I've managed to get something working by using list comprehension (someone posted an answer, but they've deleted it).
df['string'].apply(lambda row: noun_count(row),1)
which required an adjustment to my function:
def tagger_nouns(x):
list_of_lists = st.tag(x.split())
flat = [y for z in list_of_lists for y in z]
Parts_of_speech = [row[1] for row in flattened]
c = Counter(Parts_of_speech)
nouns = c['NN']+c['NNS']+c['NNP']+c['NNPS']
return nouns
I'm using the Stanford tagger, but I have a big problem with computation time, and I'm using the left 3 words model. I'm noticing that it's calling the .jar file again and again (java keeps opening and closing in the task manager) and maybe that's unavoidable, but it's really taking far too long to run. Any way I can speed it up?
I don't know what 'tagger' is but here's a simple example with a word count that ought to work more or less the same way:
f = lambda x: len(x.split())
df['num_words'] = df['string'].apply(f)
string num_words
0 'cat' 1
1 'two cats' 2
Ok so I am trying to reference one variable with another in SQL.
X= a,b,c,d (x is a string variable with a list of things in it)
Y= b ( Y is a string variable that may or may not have a vaue that appears in X)
I tried this:
Case when Y in (X) then 1 else 0 end as aa
But it doesnt work since it looks for exact matches between X and Y
also tried this:
where contains(X,#Y)
but i cant create Y globally since it is a variable that changes in each row of the table.( x also changes)
A solution in SAS would also be useful.
Thanks
Maybe like will help
select
*
from
t
where
X like ('%'+Y+'%')
or
select
case when (X like ('%'+Y+'%')) then 1 else 0 end
from
t
SQLFiddle example
In SAS I would use the INDEX function, either in a data step or proc sql. This returns the position within the string in which it finds the character(s), or zero if there is no match. Therefore a test if the value returned is greater than zero will result in a binary 1:0 output. You need to use the compress function with the variable containing the search characters as SAS pads the value with blanks.
Data step solution :
aa=index(x,compress(y))>0;
Proc Sql solution :
index(x,compress(y))>0 as aa
What is the fastest way to lookup the index of a value in sorted vector in MATLAB?
That is, is there a fast find(vector == myNumber, 1, 'first') for when vector is sorted?
I have a large matrix (200,000 x 4) of locations each with a unique integer ID recorded in the first column. I want to find the right the location of a known ID but thousands of searches can take me a little bit to find.
If you use ismembc2, the loc output should give you what you need. See this for more details:
http://www.mathworks.com/support/solutions/en/data/1-9NIE1N/index.html?product=ML&solution=1-9NIE1N
There are a number of submissions for this on FEX: http://www.mathworks.com/matlabcentral/fileexchange/?term=binary+search+vector
I do not know if it is faster but you may want to try
result=vector(vector(:,1)==myNumber,:)
result will contain the 4 elements row for which vector first column == myNumber