Enable file_uploader widget to provide information on file name and/ or type

Problem

If you allow your user to upload multiple types of files like [“csv”, “xlsx”] or [“png”, “jpg”] then you have to develop some algorithm to determine the type of file your self. You don’t know the file name either.

Solution

Return the filename together with the file object either by adding it as a name attribute to the BytesIO or StringIO object. Or by returning a tuple of (file, name).

Additional context

I experienced this problem when developing the file_uploader example for awesome-streamlit.org.

"""Streamlit v. 0.52 ships with a first version of a **file uploader** widget. You can find the
**documentation**
[here](https://streamlit.io/docs/api.html?highlight=file%20upload#streamlit.file_uploader).

For reference I've implemented an example of file upload here. It's available in the gallery at
[awesome-streamlit.org](https://awesome-streamlit.org).
"""
from enum import Enum
from io import BytesIO, StringIO
from typing import Union

import pandas as pd
import streamlit as st

STYLE = """
<style>
img {
    max-width: 100%;
}
</style>
"""

FILE_TYPES = ["csv", "py", "png", "jpg"]


class FileType(Enum):
    """Used to distinguish between file types"""

    IMAGE = "Image"
    CSV = "csv"
    PYTHON = "Python"


def get_file_type(file: Union[BytesIO, StringIO]) -> FileType:
    """The file uploader widget does not provide information on the type of file uploaded so we have
    to guess using rules or ML

    I've implemented rules for now :-)

    Arguments:
        file {Union[BytesIO, StringIO]} -- The file uploaded

    Returns:
        FileType -- A best guess of the file type
    """

    if isinstance(file, BytesIO):
        return FileType.IMAGE
    content = file.getvalue()
    if (
        content.startswith('"""')
        or "import" in content
        or "from " in content
        or "def " in content
        or "class " in content
        or "print(" in content
    ):
        return FileType.PYTHON

    return FileType.CSV


def main():
    """Run this function to display the Streamlit app"""
    st.info(__doc__)
    st.markdown(STYLE, unsafe_allow_html=True)

    file = st.file_uploader("Upload file", type=FILE_TYPES)
    show_file = st.empty()
    if not file:
        show_file.info("Please upload a file of type: " + ", ".join(FILE_TYPES))
        return

    file_type = get_file_type(file)
    if file_type == FileType.IMAGE:
        show_file.image(file)
    elif file_type == FileType.PYTHON:
        st.code(file.getvalue())
    else:
        data = pd.read_csv(file)
        st.dataframe(data.head(10))

    file.close()


main()

[svombergen]

Agreed that we need some info on the uploaded file!

For just uploading a df, as a workaround, im using a simple try/except and let it throw the error in the last try:

def try_read_df(f):
    try:
        return pd.read_csv(f)
    except:
        return pd.read_excel(f)
    
if result:
    df = try_read_df(result)
    st.dataframe(df)

[karriebear]

We’re currently working on it and targeting early Q4!