# integer

## pd.read_csv by default treats integers like floats

As root mentioned in the comments, this is a limitation of Pandas (and Numpy). NaN is a float and the empty values you have in your CSV are NaN. This is listed in the gotchas of pandas as well. You can work around this in a few ways. For the examples below I used the … Read more

## Why does the Java compiler not like primitive int as type for values in HashMap?

It’s fine with Integer, but not okay with int – Java generics only work with reference types, basically 🙁 Try this – although be aware it will box everything: HashMap<String,Integer> userName2ind = new HashMap<String,Integer>(); for (int i=0; i<=players.length; i++) { userName2ind.put(orderedUserNames[i],i+1); }

## Whats the size of an SQL Int(N)?

It depends on the database. MySQL has an extension where INT(N) means an INT with a display width of 4 decimal digits. This information is maintained in the metadata. The INT itself is still 4 bytes, and values 10000 and greater can be stored (and probably displayed, but this depends how the application uses the … Read more

## What’s the right way to divide two Int values to obtain a Float?

You have to convert the operands to floats first and then divide, otherwise you’ll perform an integer division (no decimal places). Laconic solution (requires Data.Function) foo = (/) `on` fromIntegral which is short for foo a b = (fromIntegral a) / (fromIntegral b) with foo :: Int -> Int -> Float

## Convert a string containing a hexadecimal value starting with “0x” to an integer or long

int value = (int)new System.ComponentModel.Int32Converter().ConvertFromString(“0x310530”);

## How to prevent integer overflow in Java code? [duplicate]

One way to check for an overflow is to have the operands promoted to a larger type (of double the original operand bit length) then perform the operation, and then see if the resulting value is too large for the original type, e.g. int sum(int a, int b) { long r = (long)a + b; … Read more

## Clean, efficient algorithm for wrapping integers in C++

The sign of a % b is only defined if a and b are both non-negative. int Wrap(int kX, int const kLowerBound, int const kUpperBound) { int range_size = kUpperBound – kLowerBound + 1; if (kX < kLowerBound) kX += range_size * ((kLowerBound – kX) / range_size + 1); return kLowerBound + (kX – kLowerBound) … Read more