i am scraping multiple websites so i am using one function for each website script, so each function returns 4 values, i want to print them in dataframe and write them in csv but i am facing this problem, i may be asking something too odd or basic but please help
Either i will have to write whole script in one block and that will look very nasty to handle so if i could find a way around, this is just a sample of problem i am facing..
def a1(x):
z=x+1
r = x+2
print(z, r)
def a2(x):
y=x+4
t=x+3
print(y, t)
x = 2
a1(x)
a2(x)
3 4
6 5
data = pd.Dataframe({'first' : [z],
'second' : [r],
'third' : [y],
'fourth' : [t]
})`
data
*error 'z' is not defined*
You may find it convenient to write functions that return a list of dicts.
For example:
rows = [dict(a=1, b=2, c=3),
dict(a=4, b=5, c=6)]
df = pd.DataFrame(rows)
The variables are only defined in the local scope of your functions, you'd either need to declare them globally or - the better way - return them so you can use them outside of the function by assigning the return values to new variables
import pandas as pd
def a1(x):
z = x+1
r = x+2
return (z, r)
def a2(x):
y = x+4
t = x+3
return (y, t)
x = 2
z, r = a1(x)
y, t = a2(x)
data = pd.DataFrame({'first' : [z],
'second' : [r],
'third' : [y],
'fourth' : [t]
})
Related
I am parsing some code as below. I get a list of ast.node. Now, I want to know which nodes are linked to which node. Is it possible to do that?
Example in below code: 1st assign is not connected (f = lambda..) to any other assign. But 2nd assign and FunctioDef are connected. Is it possible to infer this from some node attribute?
Assuming of course there is no other definition or code outside this.
code = '''
f = lambda x : x
factor = 2
def f(arg):
if arg == 0:
return 1
return factor*arg*f(arg-1)
'''
parsed = ast.parse(code)
>>> parsed.body
[<_ast.Assign at 0x7f9c81f0e670>,
<_ast.Assign at 0x7f9c92042070>,
<_ast.FunctionDef at 0x7f9c920421f0>]
From this
import sympy as sp
x,y,z = sp.symbols("x y z")
sp.Ep(x,y/z)
To this
#varibles = array
#equation = ????
def solver(variables,equation):
#Looping through variables array and converting variables to sympy objects
for var in variables:
var = sp.symbols(var)
#Generate sympy Equation
equation = sp.Ep(equation)
variables = [x,y,z]
equation = x,y/z #invalid code
solver(variables,equation)
I'm creating a function that is able to take in an equation with x amount of variables and x-1 number of values then solve for the missing variable symbolically then return a numerical answer using the values provided.
I only included a small portion of code where I'm having trouble understanding how to pass through an equation. Any solutions or pointers would be greatly appericated. Thanks.
There are several layers of potential confusion here concerning Python variables and SymPy objects (Symbols) used for variables.
Here is an example of what you are saying:
# 3 variables
syms = x, y, z = var('x:z')
# 2 values
vals = {x:1, y:2}
# an equations
eq = Eq(x, y/z)
# solve for the missing value symbolically
missing = set(syms) - set(vals) # == {z}
solve(eq, missing)
[y/x]
# solve for the missing value after substituting in the known values
solve(eq.subs(vals))
[2]
You could make a solver to accept an equation and then specified values and figure out the missing one and return that value by doing something like this:
>>> def solver(eq, **vals):
... from sympy.core.containers import Dict
... from sympy.solvers.solvers import solve
... free = eq.free_symbols
... vals = Dict(vals)
... x = free - set(vals)
... if len(x) != 1:
... raise ValueError('specify all but one of the values for %s' % free)
... x = x.pop()
... return solve(eq.subs(vals), x, dict=True)
...
>>> solver(eq, x=1, z=2)
[{y: 2}]
Does that give you some ideas of how to continue?
Cleaning tweet datasets by removing annoying character in bytecode (exp : \xf0\x9f\x99\x82)
Here's the code without using function :
b = data_tweet['Tweet']
b.head()
for i in b:
x = i.encode('utf=8')
y = x.decode('unicode-escape')
print(y)
It worked. The character became : 🙄, 🥰, etc.
But when I implemented it using function, in order to convert it in csv file. it failed. The byte character stays the same (exp : \xf0\x9f\x99\x82)
Here's the code :
def convert(text):
for i in text:
x = i.encode('utf=8')
y = x.decode('unicode-escape')
return text
convert(data_tweet['Tweet'])
Does anyone know why?
Problem is that you actually didn't assign the result to data_tweet['Tweet']. You can use apply() on Series.
def convert(text):
x = text.encode('utf=8')
y = x.decode('unicode-escape')
return y
data_tweet['Tweet'] = data_tweet['Tweet'].apply(convert)
Or
data_tweet['Tweet'] = data_tweet['Tweet'].apply(lambda text: text.encode('utf=8').decode('unicode-escape'))
I am running Python 3.6.2 on Windows 10 and was learning about the zip() function.
I wanted to print part of the object returned by the zip() function.
Here is my code, without the troublesome print statement:
a = ("John", "Charles", "Mike")
b = ("Jenny", "Christy", "Monica", "Vicky")
x = zip(a, b)
tup = tuple(x)
print(tup)
print(type(tup))
print(len(tup))
print(tup[1])
Here is my code with the troublesome print statement:
a = ("John", "Charles", "Mike")
b = ("Jenny", "Christy", "Monica", "Vicky")
x = zip(a, b)
print(tuple(x)[1])
tup = tuple(x)
print(tup)
print(type(tup))
print(len(tup))
print(tup[1])
The print(tuple(x)[1]) statement appears to change the tuple 'tup' into a zero-length one and causes the print(tup[1]) to fail later in the code!
In this line, you create an iterator:
x = zip(a, b)
Within the print statement, you convert the iterator to a tuple. This tuple has 3 elements. This exhausts the iterator and anytime you call it afterwards, it will return no further elements.
Therefore, upon your creation of tup, your iterator does not return an element. Hence, you have a tuple with length 0. And of course, this will raise an exception when you try to access the element with index 1.
For testing, consider this:
a = ("John", "Charles", "Mike")
b = ("Jenny", "Christy", "Monica", "Vicky")
x = zip(a, b)
tup1 = tuple(x)
tup2 = tuple(x)
print(tup1)
print(tup2)
It will give you the following result:
(('John', 'Jenny'), ('Charles', 'Christy'), ('Mike', 'Monica'))
()
This is basically what you do when creating a tuple out of an iterator twice.
I am trying to call a variable from one function into another by using the command return, without success. This is the example code I have:
def G():
x = 2
y = 3
g = x*y
return g
def H():
r = 2*G(g)
print(r)
return r
H()
When I run the code i receive the following error NameError: name 'g' is not defined
Thanks in advance!
Your function def G(): returns a variable. Therefore, when you call it, you assign a new variable for the returned variable.
Therefore you could use the following code:
def H():
G = G()
r = 2*G
print (r)
You don't need to give this statement:
return r
While you've accepted the answer above, I'd like to take the time to help you learn and clean up your code.
NameError: name 'g' is not defined
You're getting this error because g is a local variable of the function G()
Clean Version:
def multiple_two_numbers():
"""
Multiplies two numbers
Args:
none
Returns:
product : the result of multiplying two numbers
"""
x = 2
y = 3
product = x*y
return product
def main():
result = multiple_two_numbers()
answer = 2 * result
print(answer)
if __name__ == "__main__":
# execute only if run as a script
main()
Problems with your code:
Have clear variable and method names. g and G can be quiet confusing to the reader.
Your not using the if __name__ == "__main__":
Your return in H() unnecessary as well as the H() function.
Use docstrings to help make your code more readable.
Questions from the comments:
I have one question what if I had two or more variables in the first
function but I only want to call one of them
Your function can have as many variables as you want. If you want to return more than one variable you can use a dictionary(key,value) List, or Tuple. It all depends on your requirements.
Is it necessary to give different names, a and b, to the new
variables or can I use the same x and g?
Absolutely! Declaring another variable called x or y will cause the previous declaration to be overwritten. This could make it hard to debug and you and readers of your code will be frustrated.