I have xpaths as follow:
/html/body/div[1]/table[3]/tbody/tr[1]/td[3]/a
/html/body/div[1]/table[3]/tbody/tr[2]/td[3]/a
/html/body/div[1]/table[3]/tbody/tr[3]/td[3]/a
/html/body/div[1]/table[3]/tbody/tr[4]/td[3]/a
As you can see, the tr[] values are changing. I want iterate over these values.
Below is the code I have used
search_input = driver.find_elements_by_xpath('/html/body/div[1]/table[3]/tbody/tr[3]/td[3]/a')
Please let me know How can I iterate over them.
This may not be the exact solution you are looking for but this is the idea.
tableRows = driver.find_elements_by_xpath("/html/body/div[1]/table[3]/tbody/tr")
for e in tableRows:
e.find_element_by_xpath(".//td[3]/a")
if you want all third td for every row, use this:
search_input = driver.find_elements_by_xpath('/html/body/div[1]/table[3]/tbody/tr/td[3]/a')
If you want only the first 3 rows use this:
search_input = driver.find_elements_by_xpath('/html/body/div[1]/table[3]/tbody/tr[position() < 4]/td[3]/a')
For looping through tds look I.e. this answer
Another alternative, assuming 4 elements:
for elem in range(1,5):
element = f"/html/body/div[1]/table[3]/tbody/tr[{elem}]/td[3]/a"
#e = driver.find_element_by_xpath(element)
#e.click()
print(element)
Prints:
/html/body/div[1]/table[3]/tbody/tr[1]/td[3]/a
/html/body/div[1]/table[3]/tbody/tr[2]/td[3]/a
/html/body/div[1]/table[3]/tbody/tr[3]/td[3]/a
/html/body/div[1]/table[3]/tbody/tr[4]/td[3]/a
You could do whatever you wanted with the elements in the loop, I just printed to show the value
Option 1: Fixed Column number like 3 needs to be be iterated:
rows = len(self.driver.find_elements_by_xpath("/html/body/div[1]/table[3]/tbody/tr"))
for row in range(1, (rows + 1)):
local_xpath = ""/html/body/div[1]/table[3]/tbody/tr[" + str(row) + "]/td[3]"
# do something with element
# cell_text = self.driver.find_element_by_xpath(local_xpath ).text
Option 2: Both Row and Col needs to be iterated:
rows = len(self.driver.find_elements_by_xpath("/html/body/div[1]/table[3]/tbody/tr"))
columns = len(self.driver.find_elements_by_xpath("/html/body/div[1]/table[3]/tbody/tr/td"))
for row in range(1, (rows + 1)):
for column in range(1, (columns + 1)):
local_xpath = ""/html/body/div[1]/table[3]/tbody/tr[" + str(row) + "]/td[" + str(column) + "]"
# do something with element
# cell_text = self.driver.find_element_by_xpath(local_xpath ).text
Related
there is a list which contains integer values.
list=[1,2,3,.....]
then I use np.random.choice function to select a random element and add it to the a existing dataframe column, please refer to below code
df.message = df.message.astype(str) + "rowNumber=" + '"' + str(np.random.choice(list)) + '"'
But the element selected by np.random.choice and appended to the message column are always the same for all message row.
What is issue here?
Expected result is that the selected element from the list is not the same.
Pass to np.random.choice with parameter size and convert values to strings:
df = pd.DataFrame(
{'message' : ['aa','bb','cc']})
L = [1,2,3,4,5]
df.message = (df.message.astype(str) + "rowNumber=" + '"' +
np.random.choice(L, size=len(df)).astype(str) + '"')
print (df)
message
0 aarowNumber="4"
1 bbrowNumber="2"
2 ccrowNumber="5"
i am extracting selected pages from a pdf file. and want to assign dataframe name based on the pages extracted:
file = "abc"
selected_pages = ['10','11'] #can be any combination eg ['6','14','20]
for i in selected_pages():
df{str(i)} = read_pdf(path + file + ".pdf",encoding = 'ISO-8859-1', stream = True,area = [100,10,740,950],pages= (i), index = False)
print (df{str(i)} )
The idea, ultimately, as in above example, is to have dataframes: df10, df11. I have tried "df" + str(i), "df" & str(i) & df{str(i)}. however all are giving error msg: SyntaxError: invalid syntax
Or any better way of doing it is most welcome. thanks
This is where a dictionary would be a much better option.
Also note the error you have at the start of the loop. selected_pages is a list, so you can't do selected_pages().
file = "abc"
selected_pages = ['10','11'] #can be any combination eg ['6','14','20]
df = {}
for i in selected_pages:
df[i] = read_pdf(path + file + ".pdf",encoding = 'ISO-8859-1', stream = True, area = [100,10,740,950], pages= (i), index = False)
i = int(i) - 1 # this will bring it to 10
dfB = df[str(i)]
#select row number to drop: 0:4
dfB.drop(dfB.index[0:4],axis =0, inplace = True)
dfB.columns = ['col1','col2','col3','col4','col5']
I am supposed to "Write a program that uses two nested while loops to print off the rows and columns of a 3x3 grid (numbered 1 to 3), excluding the cells along the diagonal (i.e., where the row and column have the same value)."
I tried printing row, col = 1, 1 by adding one to each iteration.
row, col = 1, 1
while row != 3 and col != 3:
row += 1
col += 1
print (row, col)
Result should look like this:
1 2
1 3
2 1
vertically 12 on top, 13 in middle and 21 in the bottom.
A nested while loop generally means a second inside the first. I don't know what language you're using, but you should consider something more like:
while row <= 3:
while col <= 3:
print (row, col)
col++
col = 1
row++
This doesn't do the diagonal checks, but it demonstrates the nested loop idea:
The outer loop begins with row = 1 and col = 1.
The inner loop begins.
The inner loop counts columns 1-3, then exits;
The outer loop resets col to 1 and increments row
The next iteration of the outer loop begins with the new row.
Steps 2-5 repeat until until row = 4, at which point the outer loop exits and you're done.
Here's an example of this sort of thing in javascript, including the diagonal check:
let row = 1;
let col = 1;
// do this block until row > 3
while (row <= 3) {
// declare a new array to collect this row's output
let output = [];
// do this block until col > 3
while (col <= 3) {
// add the column or '-' to the end of the array
output.push( row === col ? '-' : col );
// increment the column
col++;
}
// row finished. emit the collected
// row with a space between the numbers
console.log(output.join(' '));
// reset column to 1
col = 1;
// do the next row
row++;
}
I need to use two Loops and the easy part is to count how many times does a "submodul" repeats in a defined and known range ("B3","B18"), this means the quantity of elements each submodul has. The difficult part comes when trying to count how many times does a "position" repeats for each different "submodul", this is because the amount of elements of each "submodul" is different so I have to adjust a range in a especial Loop to calculate how many times does a specific element (=Position) repeats within a "submodul".
The specific part that I need help with is the following:
positionrepetition = Application.CountIf(Worksheets("Sheet2").range("cells(3 + x + y - 1, 3)", "cells(3 + x + y - 1 + submodulrepetition,3"), position)
If I can manage to write it in a correct format I believe it will work. The problem is that normally I only use the range function when I know that the range is fixed or known, it doesn´t have to be calculated. I normally write for example: Range("A1","F10").Select
As you can see this is a fixed range, so I imagined that instead of using the format of Range("A1", "F10") I could use the range function with the arguments ("Cells(1,1)","Cells(10,6)"). Please correct me if I´m wrong.
Here is the rest of the code for the Loop.
For x = 0 To numberofparts
If Cells(3 + x, 18) = "1" Then
submodul = Cells(3 + x, 2).Value
submodulrepetition = Application.CountIf(Worksheets("Sheet2").range("B3", "B18"), submodul)
For y = 1 To submodulrepetition
position = Cells(3 + x + y - 1, 3).Value
positionrepetition = Application.CountIf(Worksheets("Sheet2").range("cells(3 + x + y - 1, 3)", "cells(3 + x + y - 1 + submodulrepetition,3"), position)
Next
Else
End If
x = x + submodulrepetition - 1
Next
To explain a little more, all data is gathered from Excel Sheets:
-All Information is gathered from a Excel sheet
-The "submodules" are in column B and they are arranged in numerical order. Every submodul repeats in this column as many elements it has.
-The "positions" (elements of the submodules) are in column C and can also repeat in the same column and even in other "Submodul"s.
All help will be appreciated and I thank you in advance.
Alejandro Farina
Change your line:
positionrepetition = Application.CountIf(Worksheets("Sheet2").Range("cells(3 + x + y - 1, 3)", "cells(3 + x + y - 1 + submodulrepetition,3"), Position)
With :
positionrepetition = Application.CountIf(Worksheets("Sheet2").Range(Cells(3 + x + y - 1, 3), Cells(3 + x + y - 1 + submodulrepetition, 3), Position))
If the Range is going to Change by Column/Row use the following code to get the end of column or row:
Dim GetColEnd, GetRowEnd As Integer
GetColEnd = Sheets("Sheet_Name").Cells(1, .Columns.Count).End(xlToLeft).Column
GetRowEnd = Sheets("Sheet_Name").Cells(Rows.Count, 1).End(xlUp).Row
Use the GetColEnd GetRowEnd in your Range function for flexible Column\Row for example as follows:
Sheets("Sheet_Name").Range(Cells(1,1),Cells(GetRowEnd,GetColEnd)
What is the fastest way to iterate over all elements in a 3D NumPy array? If array.shape = (r,c,z), there must be something faster than this:
x = np.asarray(range(12)).reshape((1,4,3))
#function that sums nearest neighbor values
x = np.asarray(range(12)).reshape((1, 4,3))
#e is my element location, d is the distance
def nn(arr, e, d=1):
d = e[0]
r = e[1]
c = e[2]
return sum(arr[d,r-1,c-1:c+2]) + sum(arr[d,r+1, c-1:c+2]) + sum(arr[d,r,c-1]) + sum(arr[d,r,c+1])
Instead of creating a nested for loop like the one below to create my values of e to run the function nn for each pixel :
for dim in range(z):
for row in range(r):
for col in range(c):
e = (dim, row, col)
I'd like to vectorize my nn function in a way that extracts location information for each element (e = (0,1,1) for example) and iterates over ALL elements in my matrix without having to manually input each locational value of e OR creating a messy nested for loop. I'm not sure how to apply np.vectorize to this problem. Thanks!
It is easy to vectorize over the d dimension:
def nn(arr, e):
r,c = e # (e[0],e[1])
return np.sum(arr[:,r-1,c-1:c+2],axis=2) + np.sum(arr[:,r+1,c-1:c+2],axis=2) +
np.sum(arr[:,r,c-1],axis=?) + np.sum(arr[:,r,c+1],axis=?)
now just iterate over the row and col dimensions, returning a vector, that is assigned to the appropriate slot in x.
for row in <correct range>:
for col in <correct range>:
x[:,row,col] = nn(data, (row,col))
The next step is to make
rows = [:,None]
cols =
arr[:,rows-1,cols+2] + arr[:,rows,cols+2] etc.
This kind of problem has come up many times, with various descriptions - convolution, smoothing, filtering etc.
We could do some searches to find the best, or it you prefer, we could guide you through the steps.
Converting a nested loop calculation to Numpy for speedup
is a question similar to yours. There's only 2 levels of looping, and sum expression is different, but I think it has the same issues:
for h in xrange(1, height-1):
for w in xrange(1, width-1):
new_gr[h][w] = gr[h][w] + gr[h][w-1] + gr[h-1][w] +
t * gr[h+1][w-1]-2 * (gr[h][w-1] + t * gr[h-1][w])
Here's what I ended up doing. Since I'm returning the xv vector and slipping it in to the larger 3D array lag, this should speed up the process, right? data is my input dataset.
def nn3d(arr, e):
r,c = e
n = np.copy(arr[:,r-1:r+2,c-1:c+2])
n[:,1,1] = 0
n3d = np.ma.masked_where(n == nodata, n)
xv = np.zeros(arr.shape[0])
for d in range(arr.shape[0]):
if np.ma.count(n3d[d,:,:]) < 2:
element = nodata
else:
element = np.sum(n3d[d,:,:])/(np.ma.count(n3d[d,:,:])-1)
xv[d] = element
return xv
lag = np.zeros(shape = data.shape)
for r in range(1,data.shape[1]-1): #boundary effects
for c in range(1,data.shape[2]-1):
lag[:,r,c] = nn3d(data,(r,c))
What you are looking for is probably array.nditer:
a = np.arange(6).reshape(2,3)
for x in np.nditer(a):
print(x, end=' ')
which prints
0 1 2 3 4 5