Pandas: Location of a row with error

The error you are seeing might be due to the value(s) in the x column being strings:

In [15]: df = pd.DataFrame({'x':['1.0692e+06']})
In [16]: df['x'].astype('int')
ValueError: invalid literal for long() with base 10: '1.0692e+06'

Ideally, the problem can be avoided by making sure the values stored in the DataFrame are already ints not strings when the DataFrame is built. How to do that depends of course on how you are building the DataFrame.

After the fact, the DataFrame could be fixed using applymap:

import ast
df = df.applymap(ast.literal_eval).astype('int')

but calling ast.literal_eval on each value in the DataFrame could be slow, which is why fixing the problem from the beginning is the best alternative.

---

Usually you could drop to a debugger when an exception is raised to inspect the problematic value of row.

However, in this case the exception is happening inside the call to astype, which is a thin wrapper around C-compiled code. The C-compiled code is doing the looping through the values in df['x'], so the Python debugger is not helpful here -- it won't allow you to introspect on what value the exception is being raised from within the C-compiled code.

There are many important parts of Pandas and NumPy written in C, C++, Cython or Fortran, and the Python debugger will not take you inside those non-Python pieces of code where the fast loops are handled.

So instead I would revert to a low-brow solution: iterate through the values in a Python loop and use try...except to catch the first error:

df = pd.DataFrame({'x':['1.0692e+06']})
for i, item in enumerate(df['x']):
   try:
      int(item)
   except ValueError:
      print('ERROR at index {}: {!r}'.format(i, item))

yields

ERROR at index 0: '1.0692e+06'

I hit the same problem, and as I have a big input file (3 million rows), enumerating all rows will take a long time. Therefore I wrote a binary-search to locate the offending row.

import pandas as pd
import sys

def binarySearch(df, l, r, func):
    while l <= r:
        mid = l + (r - l) // 2;

        result = func(df, mid, mid+1)
        if result:
            # Check if we hit exception at mid
            return mid, result

        result = func(df, l, mid)
        if result is None:
            # If no exception at left, ignore left half
            l = mid + 1
        else:
            r = mid - 1

    # If we reach here, then the element was not present
    return -1

def check(df, start, end):
    result = None

    try:
        # In my case, I want to find out which row cause this failure
        df.iloc[start:end].uid.astype(int)
    except Exception as e:
        result = str(e)

    return result

df = pd.read_csv(sys.argv[1])

index, result = binarySearch(df, 0, len(df), check)
print("index: {}".format(index))
print(result)