Could someone explain why I can not use int() to convert an integer number represented in string-scientific notation into a python int?
For example this does not work:
print int('1e1')
But this does:
print int(float('1e1'))
print int(1e1) # Works
Why does int not recognise the string as an integer? Surely its as simple as checking the sign of the exponent?
Behind the scenes a scientific number notation is always represented as a float internally. The reason is the varying number range as an integer only maps to a fixed value range, let's say 2^32 values. The scientific representation is similar to the floating representation with significant and exponent. Further details you can lookup in https://en.wikipedia.org/wiki/Floating_point.
You cannot cast a scientific number representation as string to integer directly.
print int(1e1) # Works
Works because 1e1 as a number is already a float.
>>> type(1e1)
<type 'float'>
Back to your question: We want to get an integer from float or scientific string. Details: https://docs.python.org/2/reference/lexical_analysis.html#integers
>>> int("13.37")
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: '13.37'
For float or scientific representations you have to use the intermediate step over float.
int(). –
Thunderous int('13.37'). –
Terrific float. –
Terrific int with scientific representation. –
Terrific >>> type(int(1e100)) <type 'long'> docs.python.org/2/reference/lexical_analysis.html#integers –
Terrific Very Simple Solution
print(int(float(1e1)))
Steps:- 1- First you convert Scientific value to float. 2- Convert that float value to int . 3- Great you are able to get finally int data type.
isinstance(1e1, float) == True. Just int(1e1) is enough, which makes this answer a duplicate. –
Gurge Because in Python (at least in 2.x since I do not use Python 3.x), If you input a string, then python will try to parse it to base 10 int() behaves differently on strings and numeric values.int
int ("077")
>> 77
But if you input a valid numeric value, then python will interpret it according to its base and type and convert it to base 10 int. then python will first interperet 077 as base 8 and convert it to base 10 then int() will jsut display it.
int (077) # Leading 0 defines a base 8 number.
>> 63
077
>> 63
So, int('1e1') will try to parse 1e1 as a base 10 string and will throw ValueError. But 1e1 is a numeric value (mathematical expression):
1e1
>> 10.0
So So Python will first interpret 1e1 since it was a numric value and evaluate int will handle it as a numeric value and handle it as though, converting it to float(10.0) and then parse it to int.10.0 and int() will convert it to integer.
So calling int() with a string value, you must be sure that string is a valid base 10 integer value.
int()) doesn't really have anything to do with Python 2 vs 3... –
Thunderous int(float(1e+001)) will work.
Whereas like what others had mention 1e1 is already a float.
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int()doesn't work on, say,'1.0': it is intended for integers. – Thunderousint('hello world')? It fails for the exact same reason thatint('1e1')does -int()parses integer strings, like it says on the tin. – Thunderous1e1into an interpreter and watch it return10.0. If that doesn't say "float" to you, trytype(1e1)and watch it return<class 'float'>. Of course, the fact that onlyfloat(), and notint(), could parse the string'1e1'is a good indication, as well. – Thunderous