is there a way that I can search for every nth term sucha as a number ranging from 0-2500 and replacing them with zeroes?
search and replace
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In a list of numbers in sequence:[code=Python]>>> def nthzero(numList , nth):Originally posted by ironmonkey69is there a way that I can search for every nth term sucha as a number ranging from 0-2500 and replacing them with zeroes?
... outList = []
... for num in numList:
... if not num%nth:
... outList.append( 0)
... else:
... outList.append( num)
... return outList
...
>>> nthzero(range(1 ,50),7)
[1, 2, 3, 4, 5, 6, 0, 8, 9, 10, 11, 12, 13, 0, 15, 16, 17, 18, 19, 20, 0, 22, 23, 24, 25, 26, 27, 0, 29, 30, 31, 32, 33, 34, 0, 36, 37, 38, 39, 40, 41, 0, 43, 44, 45, 46, 47, 48, 0]
>>> [/code]OR any list:[code=Python]def nthzero(numList , nth):
outList = []
for i, num in enumerate(numLi st):
if not (i+1)%nth:
outList.append( 0)
else:
outList.append( num)
return outList
numList = range(6,400,12)
nth = 7
print nthzero(numList , nth)
print nthzero(['x']*30, nth)
>>> [6, 18, 30, 42, 54, 66, 0, 90, 102, 114, 126, 138, 150, 0, 174, 186, 198, 210, 222, 234, 0, 258, 270, 282, 294, 306, 318, 0, 342, 354, 366, 378, 390]
['x', 'x', 'x', 'x', 'x', 'x', 0, 'x', 'x', 'x', 'x', 'x', 'x', 0, 'x', 'x', 'x', 'x', 'x', 'x', 0, 'x', 'x', 'x', 'x', 'x', 'x', 0, 'x', 'x']
>>>[/code] -
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Your question lacks specificity. Perhaps you are wanting to work with text?Originally posted by ironmonkey69what can I do if the numbers are not separated by commas?
Maybe that text is broken up with newline character, maybe not.
You may even need help reading a text file into memory.
Please be specific in your questions.
Thank you.Comment
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You can use a for loop like this.
This will work with text or numbers. First put your objects in a list. In the case of text use 0 instead of numbers[0] and the result of len(text) for numbers[-1].Code:numbers = range(0,2500) # Make list of numbers step = 5 for num in range(numbers[0], numbers[-1], step): numbers[num] = 0 # set to zero print numbers
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