Selecting rows where a numeric column value change sign through openpyxl - python-3.x

Question

I'm learning Python and openpyxl for data analysis on a large xlsx workbook. I have a for loop that can iterate down an entire column. Here's some example data:
ROW: VALUE:
1 1
2 2
3 3
4 4
5 -4
6 -1
7 -6
8 2
9 3
10 -3
I want to print out the row in which the value changes from positive to negative, and vice versa. So in the above example, row number 5, 8, and 10 would print in the console. How can I use an if statement within a for loop to iterate through a column on openpyxl?
So far I can print all of the cells in a column:
import openpyxl
wb = openpyxl.load_workbook('ngt_log.xlsx')
sheet = wb.get_sheet_by_name('sheet1')
for i in range(1, 10508, 1): # 10508 is the length of the column
print(i, sheet.cell(row=i, column=6).value)
My idea was to just add an if statement inside of the for loop:
for i in range(1, 10508, 1): # 10508 is the length of the column
if(( i > 0 and (i+1) < 0) or (i < 0 and (i+1) > 0)):
print((i+1), sheet.cell(row=i, column=6).value)
But that doesn't work. Am I formulating the if statement correctly?

Answer 1

It looks to me as though your statement is contradicting itself
for i in range(1, 10508, 1): # 10508 is the length of the column
if(( i greater than 0 and (i+1) less than 0) or (i less than 0 and (i+1) greater than
0)):
print((i+1), sheet.cell(row=i, column=6).value)
I wrote the > and < symbols in plain English but if i is greater than 0 then i + 1 is never less than 0 and vise versa so they will never work as both cannot be true

Answer 2

You need to get the sheet.cell values first, and then do the comparisons:
end_range = 10508
for i in range(1, end_range):
current, next = sheet.cell(row=i, column=6).value, sheet.cell(row=i+1, column=6).value
if current > 0 and next < 0 or current < 0 and next > 0:
print(i+1, next)
I am pretty sure there's a sign() function in the math library, but kinda overkill. You may also want to figure out what you want to do if the values are 0.

Answer 3

You can use a flag to check for positive and negative.
ws = wb['sheet1'] # why people persist in using long deprecated syntax is beyond me
flag = None
for row in ws.iter_rows(max_row=10508, min_col=6, max_col=6):
cell = row[0]
sign = cell.value > 0 and "positive" or "negative"
if flag is not None and sign != flag:
print(cell.row)
flag = sign

Answer 4

You can write the rules to select the rows where the sign has changed and put them in a generator expression without using extra memory, like this:
pos = lambda x: x>=0
keep = lambda s, c, i, v: pos(s[c][x].value)!=pos(v.value)
gen = (x+1 for x, y in enumerate(sheet['f']) if x>0 and keep(sheet, 'f', x-1, y))
Then, when you need to know the rows where the sign has changed, you just iterate on gen as below:
for row in gen:
# here you use row