colss 0.1.12

Less verbose Mathematical exp evaluator for pandas,numpy and polar

colss

colss is a lightweight expression evaluator that evaluates math-style string expressions on NumPy, Pandas, Polars, and standard Python arrays, reducing verbosity compared to native syntax eg: $\Sigma\left(e^{\sin(a)} + \log^2(b+1) - c^3\right)$

colss.sumof("exp(sin(a)) + log(b+1)^2 - c^3")

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Installation

pip install colss

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Requirements

All arrays must be 1D, preferably float64, and C-contiguous. If you have a 2D array:

a = a.ravel()

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Usage

All functions accept a string expression. Just pass the expression — no need to register variables or pass arrays manually.

query

Evaluates an expression element-wise and returns a NumPy array.

import numpy as np
import colss
a = np.array([1.0, 2.0, 3.0], dtype=np.float64)
b = np.array([4.0, 5.0, 6.0], dtype=np.float64)
colss.query("a + b + 7")
colss.query("a > 1 ? 100 : 0")       # ternary
colss.query("sqrt(a) + sin(b)")
colss.query("a ^ 2")                  # a squared
colss.query("a ^ 2 + b ^ 0.5")       # complex expression

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mean

Returns the arithmetic mean of the evaluated expression as a scalar.

colss.mean("a")
colss.mean("a + b")
colss.mean("a ^ 2 + b")

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sumof

Returns the sum of the evaluated expression as a scalar.

colss.sumof("a")
colss.sumof("a * 2")
colss.sumof("a + b")

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prod

Returns the product of all elements of the evaluated expression as a scalar.

colss.prod("a")
colss.prod("a + 1")
colss.prod("a * b")

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median

Returns the median of the evaluated expression as a scalar. For even-length arrays, returns the average of the two middle values.

colss.median("a")
colss.median("a + b")
colss.median("a ^ 2 + b")

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sd

Returns the population standard deviation of the evaluated expression as a scalar.

colss.sd("a")
colss.sd("a + b")
colss.sd("sqrt(a) + b ^ 2")

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variance

Returns the population variance of the evaluated expression as a scalar.

colss.variance("a")
colss.variance("a + b")
colss.variance("a ^ 2 - b")

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Using with Pandas

import pandas as pd
import numpy as np
import colss
df = pd.DataFrame({
    "a": [1.0, 2.0, 3.0],
    "b": [4.0, 5.0, 6.0]
})
a = df["a"].to_numpy(dtype=np.float64)
b = df["b"].to_numpy(dtype=np.float64)
df["c"] = colss.query("a + b + 7")
m       = colss.mean("a + b")
med     = colss.median("a")
s       = colss.sd("a + b")

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Operators

Operator	Description
+ - * /	Arithmetic
^	Exponentiation — a ^ 2 raises a to the power of 2
%	Modulo
> < >= <= == !=	Comparison
and or not	Logical
condition ? a : b	Ternary

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Available Functions

abs(x)       sqrt(x)      
log(x)       log10(x)     exp(x)
sin(x)       cos(x)       tan(x)
floor(x)     ceil(x)
min(x, y)    max(x, y)

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Notes

• ^ is exponentiation, not bitwise XOR.
• All arrays used in an expression must be the same length.
• You can use both scalar and vector at the same time.
• A NaN or Inf result raises an error identifying the first offending index.