I have a pandas DataFrame that I want to upload to a new CSV file. The problem is that I don't want to save the file locally before transferring it to s3. Is there any method like to_csv for writing the dataframe to s3 directly? I am using boto3.
Here is what I have so far:
import boto3
s3 = boto3.client('s3', aws_access_key_id='key', aws_secret_access_key='secret_key')
read_file = s3.get_object(Bucket, Key)
df = pd.read_csv(read_file['Body'])
# Make alterations to DataFrame
# Then export DataFrame to CSV through direct transfer to s3
This is a more up to date answer:
import s3fs
s3 = s3fs.S3FileSystem(anon=False)
# Use 'w' for py3, 'wb' for py2
with s3.open('<bucket-name>/<filename>.csv','w') as f:
df.to_csv(f)
The problem with StringIO is that it will eat away at your memory. With this method, you are streaming the file to s3, rather than converting it to string, then writing it into s3. Holding the pandas dataframe and its string copy in memory seems very inefficient.
If you are working in an ec2 instant, you can give it an IAM role to enable writing it to s3, thus you dont need to pass in credentials directly. However, you can also connect to a bucket by passing credentials to the S3FileSystem() function. See documention:https://s3fs.readthedocs.io/en/latest/
I read a csv with two columns from bucket s3, and the content of the file csv i put in pandas dataframe.
Example:
config.json
{
"credential": {
"access_key":"xxxxxx",
"secret_key":"xxxxxx"
}
,
"s3":{
"bucket":"mybucket",
"key":"csv/user.csv"
}
}
cls_config.json
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import os
import json
class cls_config(object):
def __init__(self,filename):
self.filename = filename
def getConfig(self):
fileName = os.path.join(os.path.dirname(__file__), self.filename)
with open(fileName) as f:
config = json.load(f)
return config
cls_pandas.py
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import pandas as pd
import io
class cls_pandas(object):
def __init__(self):
pass
def read(self,stream):
df = pd.read_csv(io.StringIO(stream), sep = ",")
return df
cls_s3.py
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import boto3
import json
class cls_s3(object):
def __init__(self,access_key,secret_key):
self.s3 = boto3.client('s3', aws_access_key_id=access_key, aws_secret_access_key=secret_key)
def getObject(self,bucket,key):
read_file = self.s3.get_object(Bucket=bucket, Key=key)
body = read_file['Body'].read().decode('utf-8')
return body
test.py
#!/usr/bin/env python
# -*- coding: utf-8 -*-
from cls_config import *
from cls_s3 import *
from cls_pandas import *
class test(object):
def __init__(self):
self.conf = cls_config('config.json')
def process(self):
conf = self.conf.getConfig()
bucket = conf['s3']['bucket']
key = conf['s3']['key']
access_key = conf['credential']['access_key']
secret_key = conf['credential']['secret_key']
s3 = cls_s3(access_key,secret_key)
ob = s3.getObject(bucket,key)
pa = cls_pandas()
df = pa.read(ob)
print df
if __name__ == '__main__':
test = test()
test.process()
If you pass None as the first argument to to_csv() the data will be returned as a string. From there it's an easy step to upload that to S3 in one go.
It should also be possible to pass a StringIO object to to_csv(), but using a string will be easier.
I found a very simple solution that seems to be working :
s3 = boto3.client("s3")
s3.put_object(
Body=open("filename.csv").read(),
Bucket="your-bucket",
Key="your-key"
)
Hope that helps !
You can directly use the S3 path. I am using Pandas 0.24.1
In [1]: import pandas as pd
In [2]: df = pd.DataFrame( [ [1, 1, 1], [2, 2, 2] ], columns=['a', 'b', 'c'])
In [3]: df
Out[3]:
a b c
0 1 1 1
1 2 2 2
In [4]: df.to_csv('s3://experimental/playground/temp_csv/dummy.csv', index=False)
In [5]: pd.__version__
Out[5]: '0.24.1'
In [6]: new_df = pd.read_csv('s3://experimental/playground/temp_csv/dummy.csv')
In [7]: new_df
Out[7]:
a b c
0 1 1 1
1 2 2 2
S3 File Handling
pandas now uses s3fs for handling S3 connections. This shouldn’t break any code. However, since s3fs is not a required dependency, you will need to install it separately, like boto in prior versions of pandas. GH11915.
since you are using boto3.client(), try:
import boto3
from io import StringIO #python3
s3 = boto3.client('s3', aws_access_key_id='key', aws_secret_access_key='secret_key')
def copy_to_s3(client, df, bucket, filepath):
csv_buf = StringIO()
df.to_csv(csv_buf, header=True, index=False)
csv_buf.seek(0)
client.put_object(Bucket=bucket, Body=csv_buf.getvalue(), Key=filepath)
print(f'Copy {df.shape[0]} rows to S3 Bucket {bucket} at {filepath}, Done!')
copy_to_s3(client=s3, df=df_to_upload, bucket='abc', filepath='def/test.csv')
I like s3fs which lets you use s3 (almost) like a local filesystem.
You can do this:
import s3fs
bytes_to_write = df.to_csv(None).encode()
fs = s3fs.S3FileSystem(key=key, secret=secret)
with fs.open('s3://bucket/path/to/file.csv', 'wb') as f:
f.write(bytes_to_write)
s3fs supports only rb and wb modes of opening the file, that's why I did this bytes_to_write stuff.
I use AWS Data Wrangler. For example:
import awswrangler as wr
import pandas as pd
# read a local dataframe
df = pd.read_parquet('my_local_file.gz')
# upload to S3 bucket
wr.s3.to_parquet(df=df, path='s3://mys3bucket/file_name.gz')
The same applies to csv files. Instead of read_parquet and to_parquet, use read_csv and to_csv with the proper file extension.
I found this can be done using client also and not just resource.
from io import StringIO
import boto3
s3 = boto3.client("s3",\
region_name=region_name,\
aws_access_key_id=aws_access_key_id,\
aws_secret_access_key=aws_secret_access_key)
csv_buf = StringIO()
df.to_csv(csv_buf, header=True, index=False)
csv_buf.seek(0)
s3.put_object(Bucket=bucket, Body=csv_buf.getvalue(), Key='path/test.csv')
You can also use the AWS Data Wrangler:
import awswrangler as wr
wr.s3.to_csv(
df=df,
path="s3://...",
)
Note that it will handle multipart upload for you to make the upload faster.
Source: Stackoverflow.com