I am trying to symplify the number of parameters for some calculation using symmetry properties.
In this problem I am using the script below to generate all the linear equation needed.
from sympy import *
import sympy as sym
init_printing(use_latex='mathjax')
b1=sym.Array([[Rational(-1/2),sqrt(3)/2,0],[-sqrt(3)/2,Rational(-1/2),0],[0,0,1]])
Sigma=[]
for i in range(0,3):
for j in range(0,3):
for k in range(0,3):
for l in range(0,3):
y = symbols(('C_{%d%d%d%d}')%(i+1,j+1,k+1,l+1), commutative=True)
args = []
for m in range(0,3):
for n in range(0,3):
for o in range(0,3):
for p in range(0,3):
x= symbols(('C_{%d%d%d%d}')%(m+1,n+1,o+1,p+1), commutative=True)
M=sym.Array([x])
Sigmatotal_tmp=tensorproduct(b1[m][i],b1[n][j],b1[o][k],b1[p][l],M)
args.append(Sigmatotal_tmp[0])
Sigma.append(y-Add(*args))
solve(Sigma)
In my problem C_{mnop}=C_{opmn}=C_{nmop}=..... I thought to use the sum of the indexes to eliminate something like :
if sum(m,n,o,p)= sum(n,o,p,m):
keep m,n,o,p
but this requires to to write a lot of if statement, because I am planning to go up to 10 indexes.
is there a sneaky way?
Thanks!!
If symbols are going to compare the same based on some criteria -- like sum of indices? -- then use that criteria for the name and then you won't be creating redundant symbols. If the sum is what is important then
>>> m,n,o,p=1,2,3,4
>>> Symbol("C%s"%sum((m,n,o,p)))
C10
If it's not the sum that is the invariant but just any permutation of the digits that is used, then a binary encoding of the digits used would be better. For example, say you had subscripts 1, 2 and 5 out of a possible 10 values, 0 - 9:
>>> sub = (1,2,5)
>>> ['1' if i in sub else '0' for i in range(10)]
['0', '1', '1', '0', '0', '1', '0', '0', '0', '0']
>>> int(''.join(_), 2)
400
>>> Symbol("C%s"%_)
C400
Then C400 could be the symbol name to represent that the subscripts were 1, 2 and 5. It doesn't matter if we put the numbers in little-endian or big-ending form.
Or simplest: just sort the indices (as Oscar suggested) and use them in the name:
>>> sub = (2, 1, 5)
>>> Symbol("C_{%s}" % ','.join(str(i) for i in sorted(sub)))
C_{1,2,5}
Related
So I am practicing in hacker earth and I have to take two inputs in a single line separated by space.
The below code is what I used:
x, y = [x for x in input("Enter two value: ").split()]
It is supposed to take input that looks like '2 5'
And it is returning an error:
Execution failed
ValueError : not enough values to unpack (expected 2, got 1)
Stack Trace:
Traceback (most recent call last):
File "/tmp/165461120/user_code.py", line 13, in
x, y = [x for x in input("Enter two value: ").split()]
ValueError: not enough values to unpack (expected 2, got 1)
What I think I understood is that it is giving two values as a single string. If so how do I take separate them and convert them into integers?
To take two inputs
x, y = input("Enter two value: ").split()
This should do the trick. To convert to int you can do it seprately on the both x and y.
Or better way is to use a map function
x, y = map(int, input().split())
For your case, only typecasting is remaining, so that splitted numbers becomes int. So, just you have to add
x, y = [int(x) for x in input("Enter two value: ").split()]
Alternatively, For taking 2 inputs in single line, you can use map also
x, y = map(int, input().split())
This is happening because Hacker Earth, in almost all cases, gives its input line by line.
In almost all cases, the inputs are of the form below.
1
2 3 4
5
6 7 8
This will differ on problem by problem basis and needs to be personalized for each problem.
You are getting the error not enough values to unpack because Hacker Earth is giving only a single input, and you are expecting 2. If it had been more than 2, then the error would have been too many values to unpack.
In all probability its because you are trying to input the number of test cases, which is the first input, and a single number input, in most hacker earth problems.
Process-01 : using list comprehension
whole_line = input() # 1 2 3 4 5
strings = whole_line.split(' ') # ['1', '2', '3', '4', '5']
# remove extra space within numbers if any from list
numbers = [int(num) for num in strings if len(num)>0] # [1, 2, 3, 4, 5]
print(numbers) # [1, 2, 3, 4, 5]
Now you can write down the whole logic within one line like below
numbers = [ int(num) for num in input().split(' ') if len(num)>0 ]
print(numbers)
Process-02 : using filter and map function
whole_line = input() # 1 2 3 4 5
strings = whole_line.split(' ') # ['1', '2', '3', '4', '5']
# remove extra space within numbers if any from list
strings = list(filter(lambda x :len(x)>0 ,strings))
numbers = list(map(int,strings)) # convert each string to int
print(numbers) # [1, 2, 3, 4, 5]
Now you can write down the whole logic within one line like below
numbers = list(map(int,filter(lambda x : len(x)>0,input().split(' '))))
print(numbers)
In [20]: a,b = raw_input().split()
12 12.2
In [21]: a = int(a)
Out[21]: 12
In [22]: b = float(b)
Out[22]: 12.2
You can't do this in a one-liner (or at least not without some super duper extra hackz0r skills -- or semicolons), but python is not made for one-liners.
A string = 1 2 3 4
Program should return = [[1,2],[3,4]]
in python
I want the string to be converted into a list of every two element from string
You could go for something very simple such as:
s = "10 2 3 4 5 6 7 8"
l = []
i = 0
list_split_str = s.split() # splitting the string according to spaces
while i < len(s) - 1:
l.append([s[i], s[i + 1]])
i += 2
This should output:
[['10', '2'], ['3', '4'], ['5', '6'], ['7', '8']]
You could also do something a little more complex like this in a two-liner:
list_split = s.split() # stripping spaces from the string
l = [[a, b] for a, b in zip(list_split[0::2], list_split[1::2])]
The slice here means that the first list starts at index zero and has a step of two and so is equal to [10, 3, 5, ...]. The second means it starts at index 1 and has a step of two and so is equal to [2, 4, 6, ...]. So we iterate over the first list for the values of a and the second for those of b.
zip returns a list of tuples of the elements of each list. In this case, [('10', '2'), ('3', '4'), ('5', '6'), ...]. It allows us to group the elements of the lists two by two and iterate over them as such.
This also works on lists with odd lengths.
For example, with s = "10 2 3 4 5 6 7 ", the above code would output:
[['10', '2'], ['3', '4'], ['5', '6']]
disregarding the 7 since it doesn't have a buddy.
here is the solution if the numbers exact length is divisible by 2
def every_two_number(number_string):
num = number_string.split(' ')
templist = []
if len(num) % 2 == 0:
for i in range(0,len(num),2):
templist.append([int(num[i]),int(num[i+1])])
return templist
print(every_two_number('1 2 3 4'))
you can remove the if condition and enclosed the code in try and except if you want your string to still be convert even if the number of your list is not divisible by 2
def every_two_number(number_string):
num = number_string.split(' ')
templist = []
try:
for i in range(0,len(num),2):
templist.append([int(num[i]),int(num[i+1])])
except:
pass
return templist
print(every_two_number('1 2 3 4 5'))
Based on my little knowledge on pandas，pandas.Series.str.contains can search a specific str in pd.Series. But what if the dataframe is large and I just want to glance all kinds of str element in it before I do anything?
Example like this:
pd.DataFrame({'x1':[1,2,3,'+'],'x2':[2,'a','c','this is']})
x1 x2
0 1 2
1 2 a
2 3 c
3 + this is
I need a function to return ['+','a','c','this is']
If you are looking strictly at what are string values and performance is not a concern, then this is a very simple answer.
df.where(df.applymap(type).eq(str)).stack().tolist()
['a', 'c', '+', 'this is']
There are 2 possible ways - check numeric values saved as strings or not.
Check difference:
df = pd.DataFrame({'x1':[1,'2.78','3','+'],'x2':[2.8,'a','c','this is'], 'x3':[1,4,5,4]})
print (df)
x1 x2 x3
0 1 2.8 1
1 2.78 a 4 <-2.78 is float saved as string
2 3 c 5 <-3 is int saved as string
3 + this is 4
#flatten all values
ar = df.values.ravel()
#errors='coerce' parameter in pd.to_numeric return NaNs for non numeric
L = np.unique(ar[np.isnan(pd.to_numeric(ar, errors='coerce'))]).tolist()
print (L)
['+', 'a', 'c', 'this is']
Another solution is use custom function for check if possible convert to floats:
def is_not_float_try(str):
try:
float(str)
return False
except ValueError:
return True
s = df.stack()
L = s[s.apply(is_not_float_try)].unique().tolist()
print (L)
['a', 'c', '+', 'this is']
If need all values saved as strings use isinstance:
s = df.stack()
L = s[s.apply(lambda x: isinstance(x, str))].unique().tolist()
print (L)
['2.78', 'a', '3', 'c', '+', 'this is']
You can using str.isdigit with unstack
df[df.apply(lambda x : x.str.isdigit()).eq(0)].unstack().dropna().tolist()
Out[242]: ['+', 'a', 'c', 'this is']
Using regular expressions and set union, could try something like
>>> set.union(*[set(df[c][~df[c].str.findall('[^\d]+').isnull()].unique()) for c in df.columns])
{'+', 'a', 'c', 'this is'}
If you use a regular expression for a number in general, you could omit floating point numbers as well.
Q)After executing the code Need to print the values [1, 12, 123, 2, 23, 3, 13], but iam getting [1, 12, 123, 2, 23, 3]. I have missing the letter 13. can any one tell me the reason to overcome that error?
def get_all_substrings(string):
length = len(string)
list = []
for i in range(length):
for j in range(i,length):
list.append(string[i:j+1])
return list
values = get_all_substrings('123')
results = list(map(int, values))
print(results)
count = 0
for i in results:
if i > 1 :
if (i % 2) != 0:
count += 1
print(count)
Pretty straight forward issue in your nested for loops within get_all_substrings(), lets walk it!
You are iterating over each element of your string 123:
for i in range(length) # we know length to be 3, so range is 0, 1, 2
You then iterate each subsequent element from the current i:
for j in range(i,length)
Finally you append a string from position i to j+1 using the slice operator:
list.append(string[i:j+1])
But what exactly is happening? Well we can step through further!
The first value of i is 0, so lets skip the first for, go to the second:
for j in range(0, 3): # i.e. the whole string!
# you would eventually execute all of the following
list.append(string[0:0 + 1]) # '1'
list.append(string[0:1 + 1]) # '12'
list.append(string[0:2 + 1]) # '123'
# but wait...were is '13'???? (this is your hint!)
The next value of i is 1:
for j in range(1, 3):
# you would eventually execute all of the following
list.append(string[1:1 + 1]) # '2'
list.append(string[1:2 + 1]) # '23'
# notice how we are only grabbing values of position i or more?
Finally you get to i is 2:
for j in range(2, 3): # i.e. the whole string!
# you would eventually execute all of the following
list.append(string[2:2 + 1]) # '3'
I've shown you what is happening (as you've asked in your question), I leave it to you to devise your own solution. A couple notes:
You need to look at all index combinations from position i
Dont name objects by their type (i.e. dont name a list object list)
I would try something like this using itertools and powerset() recipe
from itertools import chain, combinations
def powerset(iterable):
s = list(iterable)
return chain.from_iterable(combinations(s, r) for r in range(len(s) + 1))
output = list(map(''.join, powerset('123')))
output.pop(0)
Here is another option, using combinations
from itertools import combinations
def get_sub_ints(raw):
return [''.join(sub) for i in range(1, len(raw) + 1) for sub in combinations(raw, i)]
if __name__ == '__main__':
print(get_sub_ints('123'))
>>> ['1', '2', '3', '12', '13', '23', '123']
I use Python 3
I want to check if all of my tested values in the nested dictionary are non 0.
So here is the simplified example dict:
d = {'a': {'1990': 10, '1991': 0, '1992': 30},
'b': {'1990': 15, '1991': 40, '1992': 0}}
and I want to test if for both dicts 'a' and 'b' the values of the keys '1990' and '1991' are not zero
for i in d:
for k in range(2):
year = 1990
year = year + k
if all((d[i][str(year)]) != 0):
print(d[i])
so it should only return b, because a['1991']=0
but this is the first time I work with the all() function and I get the error core: TypeError: 'bool' object is not iterable
the error is in the if all() line
thank you very much!
This can done a bit more generally with a list comprehension where you iterate over the items in dict d. A simple comprehension to iterate over the keys and values in our dictionary looks like this:
>>> [k for k, v in d.items()]
['a', 'b']
In the above k will contain the keys and v the values. The comprehension also has an if clause. With that you can filter out the items you don't want. So we define years = ('1990', '1991'). Now we can do another comprehension to test our year values.
To iterate over only 'a', we could do this:
>>> [d['a'][y] for y in years]
[10, 0]
>>> all([d['a'][y] for y in years])
False
Gluing the whole thing together:
>>> d={'a' :{ '1990': 10, '1991':0, '1992':30},'b':{ '1990':15, '1991':40, '1992':0}}
>>> years = ('1990', '1991')
>>> [k for k, v in d.items() if all([v[y] for y in years])]
['b']
See the python docs for more information on list comprehensions.