I have two folders each contains about 8,000 small csv files. One with an aggregated size of around 2GB and another with aggregated size of around 200GB.
These files are stored like this to better update them in a daily basis. However, when I conduct EDA, I would like them to be assigned to a single variable. For example.
path = "some random path"
df = pd.concat([pd.read_csv(f"{path}//{files}") for files in os.listdir(path)])
It would take much less time for me to read the dataset with 2GB in total size than reading it on the super computer cluster. And it is impossible to read the 200GB dataset on the local machine unless using some sort of scaling Pandas solutions. The situation does not seem to improve on the cluster even using the popular open-source tools like Dask and Modin.
Is there an effective way that enables to read those csv files effectively with given situation?
Q :"Is there an effective way that enables to read those csv files effectively ... ?"
A :Oh, sure, there is :
CSV format ( standard attempts in RFC4180 ) is not unambiguous and is not obeyed under all circumstances ( commas inside fields, header present or not ), so some caution & care is needed here. Given you are your own data curator, you shall be able to decide plausible steps for handling your own data properly.
So, the as-is state is :
# in <_folder_1_>
:::::::: # 8000 CSV-files ~ 2GB in total
||||||||||||||||||||||||||||||||||||||||||| # 8000 CSV-files ~ 200GB in total
# in <_folder_2_>
Speaking efficiency, O/S coreutils provide the best, stable, proven and most efficient (as system tool used to be since ever ) tools for the phase of merging thousands and thousands of plain CSV-files' content :
###################### if need be,
###################### use an in-place remove of all CSV-file headers first :
for F in $( ls *.csv ); do sed -i '1d' $F; done
this helps for case we cannot avoid headers on the CSV-exporter side. Works like this :
(base):~$ cat ?.csv
HEADER
1
2
3
HEADER
4
5
6
HEADER
7
8
9
(base):~$ for i in $( ls ?.csv ); do sed -i '1d' $i; done
(base):~$ cat ?.csv
1
2
3
4
5
6
7
8
9
Now, the merging phase :
###################### join
cat *.csv > __all_CSVs_JOINED.csv
Given the nature of the said file storage policy, performance can be boosted by using more processes for independent taking small files and large files separately, as defined above, having put the logic inside a pair of conversion_script_?.sh shell-scripts :
parallel --jobs 2 conversion_script_{1}.sh ::: $( seq -f "%1g" 1 2 )
As the transformation is a "just"-[CONCURRENT] flow of processing for a sake of removing the CSV-headers, but a pure-[SERIAL] ( for larger number of files, there might become interesting to use a multi-staged tree of trees - using several stages of [SERIAL]-collections of [CONCURRENT]-ly pre-processed leaves, yet for just 8000 files, not knowing the actual file-system details, the latency-masking from a just-[CONCURRENT] processing both of the directories just independently will be fine to start with )
Last but not least, the final pair of ___all_CSVs_JOINED.csv are safe to get opened using in a way, that prevents moving all disk-stored date into RAM at once ( using chunk-size-fused file-reading-iterator, avoiding RAM-spillovers by using mmaped-mode as a context manager ) :
with pandas.read_csv( "<_folder_1_>//___all_CSVs_JOINED.csv",
sep = NoDefault.no_default,
delimiter = None,
...
chunksize = SAFE_CHUNK_SIZE,
...
memory_map = True,
...
) \
as df_reader_MMAPer_CtxMGR:
...
When tweaking for ultimate performance, details matter and depend on physical hardware bottlenecks ( disk-I/O-wise, filesystem-wise, RAM-I/O-wise ), so due care may take further improvement for minimising the repetitive performed end-to-end processing times ( sometimes even turning data into a compressed/zipped form, in cases, where CPU/RAM resources permit sufficient performance advantages over limited performance of disk-I/O throughput - moving less bytes is so faster, that CPU/RAM-decompression costs are still lower, than moving 200+ [GB]s of uncompressed plain text data.
Details matter,tweak options,benchmark,tweak options,benchmark,tweak options,benchmark
would be nice to post your progress on testing the performanceend-2-end duration of strategy ... [s] AS-IS nowend-2-end duration of strategy ... [s] with parallel --jobs 2 ...end-2-end duration of strategy ... [s] with parallel --jobs 4 ...end-2-end duration of strategy ... [s] with parallel --jobs N ... + compression ... keep us posted
Related
Is it possible to do a dry run for snakemake for a single wildcard, in the order of execution?
When I call a dry run, I get the following at the bottom:
Job counts:
count jobs
1 all
1 assembly_eval
5 cat_fastq
1 createGenLogDir
5 createLogDir
5 flye
5 medaka_first
5 medaka_second
5 minimap_first
5 quast_medaka_first
5 quast_medaka_second
5 quast_racon_first
5 racon_first
5 symLinkFQ
58
This was a dry-run (flag -n). The order of jobs does not reflect the order of execution.
So I guess it would be useful to:
get the dry run commands for a single wildcard (except for the aggregate rules, obviously), after all, the only thing that differs among the commands of any of those rules is the wildcard in the input, output and param directives.
get the workflow printed in the order of execution, for enhanced visualisation.
I did not find a suitable option using snakemake -h, and I'd be looking for something that --rulegraph, does compared --dag, which is to avoid redundancy.
If there is no solution to this, or if the solution is too cumbersome, I guess I will suggest this as enhancement in their github page.
Here are some possible solutions:
You can specify a target file with the specific wildcard you want, e.g. snakemake -nq output_wc1.txt
If your wildcards are stored in a list/dataframe, limit to just the first. I frequently do this while developing, e.g. chroms = range(1,2) # was range(1, 23)
If you have a single job for each rule and dependencies are simple (A -> B -> C), the jobs should be listed in order of execution. This is not true when your workflow has concurrent or branching rules.
Have you also checked --filegraph and --summary?
I'm able to get the stats (additions, deletions, total) for each commit, however how can I get the overall #?
For example, if one MR has 30 commits, I need the net # of lines of code added\deleted which you can see in the top corner.
This # IS NOT the sum of all #'s per commit.
So, I would need an API that returns the net # of lines of code added\removed at MR level (no matter how many commits are).
For example, if I have 2 commits: 1st one adds 10 lines, and the 2nd one removes the exact same 10 lines, then the net # is 0.
Here is the scenario:
I have an MR with 30 commits.
GitLab API provides support to get the stats (lines of code added\deleted) per Commit (individually).
If I go in GitLab UI, go to the MR \ Changes, I see the # of lines added\deleted that is not the SUM of all the Commits stats that I'm getting thru API.
That's my issue.
A simpler example: let's say I have 2 commits, one adds 10 lines of code, while the 2nd commit removes the exact same 10 lines of code. Using the API, I'm getting the sum, which is 20 LOCs added. However, if I go in the GitLab UI \ Changes, it's showing me 0 (zero), which is correct; that's the net # of chgs overall. This is the inconsistency I noticed.
To do this for an MR, you would use the MR changes API and count the occurrences of lines starting with + and - in the changes[].diff fields to get the additions and deletions respectively.
Using bash with gitlab-org/gitlab-runner!3195 as an example:
GITLAB_HOST="https://gitlab.com"
PROJECT_ID="250833"
MR_ID="3195"
URL="${GITLAB_HOST}/api/v4/projects/${PROJECT_ID}/merge_requests/${MR_ID}/changes"
DIFF=$(curl ${URL} | jq -r ".changes[].diff")
ADDITIONS=$(grep -E "^\+" <<< "$DIFF")
DELETIONS=$(grep -E "^\-" <<< "$DIFF")
NUM_ADDITIONS=$(wc -l <<< "$ADDITIONS")
NUM_DELETIONS=$(wc -l <<< "$DELETIONS")
echo "${MR_ID} has ${NUM_ADDITIONS} additions and ${NUM_DELETIONS} deletions"
The output is
3195 has 9 additions and 2 deletions
This matches the UI, which also shows 9 additions and 2 deletions
This, as you can see is a representative example of your described scenario since the combined total of the individual commits in this MR are 13 additions and 6 deletions.
ch_files = Channel.fromPath("myfiles/*.csv")
ch_parameters = Channel.from(['A','B, 'C', 'D'])
ch_samplesize = Channel.from([4, 16, 128])
process makeGrid {
input:
path input_file from ch_files
each parameter from ch_parameters
each samplesize from ch_samplesize
output:
tuple path(input_file), parameter, samplesize, path("config_file.ini") into settings_grid
"""
echo "parameter=$parameter;sampleSize=$samplesize" > config_file.ini
"""
}
gives me a number_of_files * 4 * 3 grid of settings files, so I can run some script for each combination of parameters and input files.
How do I add some ID to each line of this grid? A row ID would be OK, but I would even prefer some unique 6-digit alphanumeric code without a "meaning" because the order in the table doesn't matter. I could extract out the last part of the working folder which is seemingly unique per process; but I don't think it is ideal to rely on sed and $PWD for this, and I didn't see it provided as a runtime metadata variable provider. (plus it's a bit long but OK). In a former setup I had a job ID from the LSF cluster system for this purpose, but I want this to be portable.
Every combination is not guaranteed to be unique (e.g. having parameter 'A' twice in the input channel should be valid).
To be clear, I would like this output
file1.csv A 4 pathto/config.ini 1ac5r
file1.csv A 16 pathto/config.ini 7zfge
file1.csv A 128 pathto/config.ini ztgg4
file2.csv A 4 pathto/config.ini 123js
etc.
Given the input declaration, which uses the each qualifier as an input repeater, it will be difficult to append some unique id to the grid without some refactoring to use either the combine or cross operators. If the inputs are just files or simple values (like in your example code), refactoring doesn't make much sense.
To get a unique code, the simple options are:
Like you mentioned, there's no way, unfortunately, to access the unique task hash without some hack to parse $PWD. Although, it might be possible to use BASH parameter substitution to avoid sed/awk/cut (assuming BASH is your shell of course...) you could try using: "${PWD##*/}"
You might instead prefer using ${task.index}, which is a unique index within the same task. Although the task index is not guaranteed to be unique across executions, it should be sufficient in most cases. It can also be formatted for example:
process example {
...
script:
def idx = String.format("%06d", task.index)
"""
echo "${idx}"
"""
}
Alternatively, create your own UUID. You might be able to take the first N characters but this will of course decrease the likelihood of the IDs being unique (not that there was any guarantee of that anyway). This might not really matter though for a small finite set of inputs:
process example {
...
script:
def uuid = UUID.randomUUID().toString()
"""
echo "${uuid}"
echo "${uuid.take(6)}"
echo "${uuid.takeBefore('-')}"
"""
}
I have a dataset on the server1 that I want to back up to the second server2.
Server1 (original):
zfs list -o name,used,avail,refer,creation,usedds,usedsnap,origin,compression,compressratio,refcompressratio,mounted,atime,lused storage/iscsi/webhost-old produces:
NAME USED AVAIL REFER CREATION USEDDS USEDSNAP ORIGIN COMPRESS RATIO REFRATIO MOUNTED ATIME LUSED
storage/iscsi/webhost-old 67,8G 1,87T 67,8G Út kvě 31 6:54 2016 67,8G 16K - lz4 1.00x 1.00x - - 67,4G
Sending volume to the 2nd server:
zfs send storage/iscsi/webhost-old | pv | ssh -c arcfour,aes128-gcm#openssh.com root#10.0.0.2 zfs receive -Fduv pool/bkp-storage
received 69,6GB stream in 378 seconds (189MB/sec)
Server2 zfs list produces:
NAME USED AVAIL REFER CREATION USEDDS USEDSNAP ORIGIN COMPRESS RATIO REFRATIO MOUNTED ATIME LUSED
pool/bkp-storage/iscsi/webhost-old 36,1G 3,01T 36,1G Pá pro 29 10:25 2017 36,1G 0 - lz4 1.15x 1.15x - - 28,4G
Why is there such a difference in sizes? Thanks.
From what you posted, I noticed 3 things that seemed odd:
the compressratio is 1.15x on system 2, but 1.00x on system 1
on system 2, used is 1.27x higher than logicalused
the logicalused and the number zfs receive report are ~2.3x higher on system 1 than system 2
These terms are all defined in the man page, but are still confusing to reverse-engineer explanations for in practice.
(1) could happen if you enabled compression on the source dataset after you wrote all the data to it, since ZFS doesn't rewrite the data to compress it when you enable that setting. The data sent by zfs send is uncompressed unless you use -c, but system 2 will try to compress it as it runs zfs receive if the setting is enabled on the destination dataset. If both system 1 and system 2 had the same compression settings before the data was written, they would have the same compressratio as well.
(2) can happen due to metadata written along with your data, but in this case it's too high for "normal" metadata, which accounts for 1-2% of most pools. It's probably caused by a pool-wide setting, like configuring RAID-Z, or a weird combination of striping and mirroring (like 4 stripes, but with one of them being a mirror).
For (3), I re-read the man page to try to figure it out:
logicalused
The amount of space that is "logically" consumed by this dataset and
all its descendents. See the used property. The logical space
ignores the effect of the compression and copies properties, giving a
quantity closer to the amount of data that applications see.
If you were sending a dataset (instead of a single iSCSI volume) and the send size matched system 2's logicalused value (instead of system 1's), I would guess you forgot to send some child datasets (i.e. by using zfs send -R). However, neither of those are true in this case.
I had to do some additional digging -- this blog post from 2005 might contain the explanation. If system 1 didn't have compression enabled when the data was written (like I guessed above for (1)), the function responsible for not writing zeroed-out blocks (zio_compress_data) would not be run, so you probably have a bunch of empty blocks written to disk, and accounted for in the logicalused size. However, since lz4 is configured on system 2, it would run there, and those blocks would not be counted.
I have some fairly simple Hadoop streaming jobs that look like this:
yarn jar /usr/lib/hadoop-mapreduce/hadoop-streaming-2.2.0.2.0.6.0-101.jar \
-files hdfs:///apps/local/count.pl \
-input /foo/data/bz2 \
-output /user/me/myoutput \
-mapper "cut -f4,8 -d," \
-reducer count.pl \
-combiner count.pl
The count.pl script is just a simple script that accumulates counts in a hash and prints them out at the end - the details are probably not relevant but I can post it if necessary.
The input is a directory containing 5 files encoded with bz2 compression, roughly the same size as each other, for a total of about 5GB (compressed).
When I look at the running job, it has 45 mappers, but they're all running on one node. The particular node changes from run to run, but always only one node. Therefore I'm achieving poor data locality as data is transferred over the network to this node, and probably achieving poor CPU usage too.
The entire cluster has 9 nodes, all the same basic configuration. The blocks of the data for all 5 files are spread out among the 9 nodes, as reported by the HDFS Name Node web UI.
I'm happy to share any requested info from my configuration, but this is a corporate cluster and I don't want to upload any full config files.
It looks like this previous thread [ why map task always running on a single node ] is relevant but not conclusive.
EDIT: at #jtravaglini's suggestion I tried the following variation and saw the same problem - all 45 map jobs running on a single node:
yarn jar \
/usr/lib/hadoop-mapreduce/hadoop-mapreduce-examples-2.2.0.2.0.6.0-101.jar \
wordcount /foo/data/bz2 /user/me/myoutput
At the end of the output of that task in my shell, I see:
Launched map tasks=45
Launched reduce tasks=1
Data-local map tasks=18
Rack-local map tasks=27
which is the number of data-local tasks you'd expect to see on one node just by chance alone.