How to ignore the first line of data when processing CSV data?

Question

You could use an instance of the csv module's Sniffer class to deduce the format of a CSV file and detect whether a header row is present along with the built-in next() function to skip over the first row only when necessary:

import csv

with open('all16.csv', 'r', newline='') as file:
    has_header = csv.Sniffer().has_header(file.read(1024))
    file.seek(0)  # Rewind.
    reader = csv.reader(file)
    if has_header:
        next(reader)  # Skip header row.
    column = 1
    datatype = float
    data = (datatype(row[column]) for row in reader)
    least_value = min(data)

print(least_value)

Since datatype and column are hardcoded in your example, it would be slightly faster to process the row like this:

    data = (float(row[1]) for row in reader)

Note: the code above is for Python 3.x. For Python 2.x use the following line to open the file instead of what is shown:

with open('all16.csv', 'rb') as file:

To skip the first line just call:

next(inf)

Files in Python are iterators over lines.

Borrowed from python cookbook,
A more concise template code might look like this:

import csv
with open('stocks.csv') as f:
    f_csv = csv.reader(f) 
    headers = next(f_csv) 
    for row in f_csv:
        # Process row ...

In a similar use case I had to skip annoying lines before the line with my actual column names. This solution worked nicely. Read the file first, then pass the list to csv.DictReader.

with open('all16.csv') as tmp:
    # Skip first line (if any)
    next(tmp, None)

    # {line_num: row}
    data = dict(enumerate(csv.DictReader(tmp)))