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I wish to find out the minimum value from the 3rd column of this particular data set. I have somehow used a "jugaad" method to get the minimum value by setting the variable to higher value. However i wish to know if there is a cleaner alternative to this. I am learning python and i have researched that by using numpy it's easier, however i'd prefer the method without using it.

min=10000
data=[
[514.8,526.9,512,525,27942414],
[527.9,528.5,507.15,499,19353195],
[510,522.2,504.8,520.85,21799071],
[519.95,523.2,512,515.7,18364451],
[510.4,513.85,494.25,501.85,17946821],
]
for i in range(5):
    if min>data[i][3]:
        min=data[i][3]
print(min)
Sauhard Koolwal
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  • does this help : https://stackoverflow.com/questions/51055867/numpy-matrix-column-wise-minimum-value-index – oubaydos Jan 28 '22 at 14:45
  • Does this answer your question? [Numpy matrix column wise minimum value index](https://stackoverflow.com/questions/51055867/numpy-matrix-column-wise-minimum-value-index) – oubaydos Jan 28 '22 at 14:45
  • `min(data, key=lambda x: x[3])[3]`. In your approach you can set `min=data[0][3]` as the starting value. Please avoid shadowing built-in functions with variable names. – Michael Szczesny Jan 28 '22 at 14:47
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    Third column or fourth? Remember that Python lists are base zero indexed. Also, overwriting the builtin *min* function will end in tears – DarkKnight Jan 28 '22 at 14:52
  • Third column by assuming zero indexed. – Sauhard Koolwal Jan 28 '22 at 14:53
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    In which case, take @MichaelSzczesny 's excellent answer but replace occurrences of 3 with 2 – DarkKnight Jan 28 '22 at 14:54

2 Answers2

1

You can use generator object to make it cleaner:

m = min(i[3] for i in data)

Output:

499

Also, don't use min as a variable as it is a function name.

Eli Harold
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@EliHarold's answer is great. It is very concise. It uses list comprehensions, which I found to be confusing when I first start using Python. Even now, I often prefer to use an explicit for loop.

Here is one brute force method that you could use to first extract the third column, and then use the built-in min function in Python to find the minimum of a 1D list. Python uses zero-based indexes, so if you want the third column, you need to use index 2.

data=[
[514.8,526.9,512,525,27942414],
[527.9,528.5,507.15,499,19353195],
[510,522.2,504.8,520.85,21799071],
[519.95,523.2,512,515.7,18364451],
[510.4,513.85,494.25,501.85,17946821],
]

col3 = []

for row in data:
    col3.append(row[2])

minimum = min(col3)
print(minimum)

But even your original code will work if you use an alternate form of the range function*:

range(stop)
range(start, stop[, step])

If we use the second version of the syntax in the form of range(1, step), then we can use the first row to set a minval to compare to and range through second and following rows:

data=[
[514.8,526.9,512,525,27942414],
[527.9,528.5,507.15,499,19353195],
[510,522.2,504.8,520.85,21799071],
[519.95,523.2,512,515.7,18364451],
[510.4,513.85,494.25,501.85,17946821],
]

minval = data[0][2]

for i in range(1, len(data)):
    if minval > data[i][2]:
        minval = data[i][2]

print(minval)

* OK, strictly speaking range() isn't a function, but a type that is a immutable sequence of numbers.

bfris
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