Python中如何对比CSV列表相邻元素并输出1或0

First, we need to clean your dataset to remove non-numeric entries and separator rows, then perform the adjacent value comparison. Here's a step-by-step implementation:

Step 1: Clean the Data #

Filter out separator rows (those with - and NaN values) and extract only numeric values from valid data rows:

import pandas as pd

# Your existing dataset (h111)
h111 = [['2', '1', ',', '5', '6', '9', '.', '6.1', '3'], ['2', '1', ',', 5.0, 6.0, 9.0, '.', 6.0, 3.0], ['1', '8', ',', 9.0, 5.0, 2.0, '.', 9.0, 2.0], ['-', '-', float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan')], ['1', '8', ',', 9.0, 4.0, 4.0, '.', 3.0, 6.0], ['1', '8', ',', 9.0, 4.0, 4.0, '.', 3.0, 6.0], ['1', '8', ',', 9.0, 4.0, 4.0, '.', 3.0, 6.0], ['1', '8', ',', 9.0, 4.0, 4.0, '.', 3.0, 6.0], ['-', '-', float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan')], ['1', '8', ',', 9.0, 4.0, 4.0, '.', 3.0, 6.0], ['-', '-', float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan')], ['2', '0', ',', 1.0, 5.0, 0.0, '.', 5.0, 3.0], ['2', '0', ',', 1.0, 5.0, 0.0, '.', 5.0, 3.0], ['2', '0', ',', 1.0, 5.0, 0.0, '.', 5.0, 3.0], ['2', '0', ',', 1.0, 5.0, 0.0, '.', 5.0, 3.0], ['2', '0', ',', 1.0, 6.0, 0.0, '.', 8.0, 6.0], ['2', '0', ',', 1.0, 8.0, 2.0, '.', 5.0, 9.0], ['2', '0', ',', 1.0, 7.0, 8.0, '.', 0.0, 4.0], ['2', '0', ',', 0.0, 6.0, 8.0, '.', 3.0, 2.0], ['2', '0', ',', 0.0, 5.0, 6.0, '.', 8.0, 6.0], ['2', '0', ',', 0.0, 5.0, 6.0, '.', 8.0, 6.0], ['2', '0', ',', 0.0, 5.0, 6.0, '.', 8.0, 6.0], ['2', '0', ',', 0.0, 5.0, 6.0, '.', 8.0, 6.0], ['2', '0', ',', 0.0, 8.0, 0.0, '.', 2.0, 1.0], ['2', '0', ',', 0.0, 3.0, 7.0, '.', 8.0, 2.0], ['1', '9', ',', 9.0, 6.0, 8.0, '.', 9.0, 6.0], ['1', '9', ',', 8.0, 4.0, 6.0, '.', 7.0, 0.0], ['1', '9', ',', 8.0, 4.0, 1.0, '.', 8.0, 0.0], ['-', '-', float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan')], ['1', '9', ',', 3.0, 4.0, 0.0, '.', 8.0, 4.0], ['1', '8', ',', 9.0, 1.0, 2.0, '.', 4.0, 4.0], ['1', '8', ',', 9.0, 1.0, 2.0, '.', 4.0, 4.0], ['1', '8', ',', 8.0, 8.0, 6.0, '.', 6.0, 2.0], ['1', '8', ',', 8.0, 8.0, 6.0, '.', 6.0, 2.0], ['1', '8', ',', 8.0, 4.0, 7.0, '.', 5.0, 6.0], ['-', '-', float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan')], ['1', '8', ',', 9.0, 9.0, 1.0, '.', 2.0, 2.0], ['1', '9', ',', 0.0, 0.0, 8.0, '.', 3.0, 3.0], ['-', '-', float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan')], ['-', '-', float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan')], ['-', '-', float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan')], ['-', '-', float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan')], ['1', '9', ',', 0.0, 4.0, 4.0, '.', 1.0, 0.0], ['-', '-', float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan')], ['5', '.', '4', 0.0, float('nan'), float('nan'), float('nan'), float('nan'), float('nan')], ['-', '-', float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan'), float('nan')]]

# Filter separator rows and extract numeric values
clean_rows = []
for row in h111:
    # Skip separator rows (identified by first element '-' and third element NaN)
    if row[0] == '-' and pd.isna(row[2]):
        continue
    # Extract numeric values from the row
    numeric_vals = []
    for val in row:
        try:
            # Convert value to float (handles string numbers like '2' and numeric types)
            num = float(val)
            numeric_vals.append(num)
        except (ValueError, TypeError):
            # Skip non-numeric values like ',', '.'
            continue
    if numeric_vals:
        clean_rows.append(numeric_vals)

# Flatten into a single list of numeric values
flattened_numeric = [num for row in clean_rows for num in row]

Step 2: Compare Adjacent Values #

Iterate through the cleaned list and generate the result (1 if current > next, 0 otherwise; equal values default to 0):

result = []
for i in range(len(flattened_numeric) - 1):
    current = flattened_numeric[i]
    next_val = flattened_numeric[i + 1]
    if current > next_val:
        result.append(1)
    else:
        # Covers both current < next and current == next
        result.append(0)

print(result)

Notes #

• If you want to handle equal values differently (e.g., output None or a separate flag), modify the else clause accordingly.
• If you need to compare elements row-wise (each element in a row with the corresponding element in the next row), use this alternative code instead of flattening:

row_wise_result = []
for i in range(len(clean_rows) - 1):
    row1 = clean_rows[i]
    row2 = clean_rows[i + 1]
    # Compare up to the length of the shorter row
    min_length = min(len(row1), len(row2))
    row_result = []
    for j in range(min_length):
        val1 = row1[j]
        val2 = row2[j]
        row_result.append(1 if val1 > val2 else 0)
    row_wise_result.append(row_result)

print(row_wise_result)

内容的提问来源于stack exchange，提问作者user 456354