[FIXED] How to upload a file in FastAPI and convert it into a Pandas Dataframe?

Issue

How could I convert a file into a dataframe using the following code?

@app.post("/uploadfile/")
async def create_upload_file(file: UploadFile):
    df = pd.read_csv("")
    print(df)
    return {"filename": file.filename}

Solution

Option 1

Convert the bytes into a string and then load it into an in-memory text buffer (i.e., StringIO), which can be converted into a dataframe:

from fastapi import FastAPI, File, UploadFile
import pandas as pd
from io import StringIO

app = FastAPI()

@app.post("/uploadfile/")
async def create_upload_file(file: UploadFile = File(...)):
    contents = await file.read()
    s = str(contents,'utf-8')
    data = StringIO(s) 
    df = pd.read_csv(data)
    data.close()
    return {"filename": file.filename}

Option 2

Use an in-memory bytes buffer instead (i.e., BytesIO), thus saving you the step of converting the bytes into a string:

from fastapi import FastAPI, File, UploadFile
import pandas as pd
from io import BytesIO

app = FastAPI()

@app.post("/uploadfile/")
async def create_upload_file(file: UploadFile = File(...)):
    contents = await file.read()
    data = BytesIO(contents)
    df = pd.read_csv(data)
    data.close()
    return {"filename": file.filename}

Answered By - Chris

Answer Checked By - Senaida (PHPFixing Volunteer)