Datasets:

mteb
/

arena-results

+{"tstamp": 1722266149.2701, "task_type": "retrieval", "type": "leftvote", "models": ["", ""], "ip": "", "0_conv_id": "8c04b396b11b41daa58f4645afb37357", "0_model_name": "mixedbread-ai/mxbai-embed-large-v1", "0_prompt": "do dusk balls work in max raid dens?", "0_output": [["do dusk balls work in max raid dens?", "Gaming Stackexchange\n\nQ: Do turns carry over from previous encounters with fleeing Legendary Pok\u00e9mon? I've been trying to catch Thundurus, but I don't know which is the best Pok\u00e9 Ball, the Quick ball or the Ultra ball? \nI left the wild Thundurus with 1 hp and asleep, but I don't know if it counts as one turn on the new encounters or if my previous turns carry over. I am using both types of Pok\u00e9 Balls at the moment, and it does not look like any are good. >.<\n\nA: Despite the fact that you may have been pecking away at wandering pokemon for many, many turns, whenever you meet them, it starts a new battle, from turn 1.\nThis means that the Quick Ball has twice the capture rate (x4) than that of the Ultra Ball (x2).\n"]], "0_corpus": "stackexchange", "1_conv_id": "ecb249a3187244458381cbb0af7b2bb2", "1_model_name": "sentence-transformers/all-MiniLM-L6-v2", "1_prompt": "do dusk balls work in max raid dens?", "1_output": [["do dusk balls work in max raid dens?", "Stackoverflow Stackexchange\n\nQ: AWS-SDK: Query parameter in listobjects of S3 I want to fetch objects after a particular date. Using the AWS CLI I can list objects using below command: \naws s3api list-objects-v2 --bucket \"bucket1\" --prefix \"file-\" --query \"(Contents[?LastModified>'2019-02-06T05:34:12.000Z'])[0]\"\nBut I want to do it from my code so please let me know how can I filter objects usin NPM AWS-SDK.\nNote: I can do it using exec or spawn but for this I have to configure profile using CLI which will create credential file on local so I dont want to do this.\n\nA: Use the AWS SDK for node.js. Call the listObjectsV2 method and then use jmespath.js in the callback method to filter the output of the API call. This is the equivalent of what the AWS CLI does via the --query param.\nSomething like this (untested)\nvar params = {\n  Bucket: \"bucket1\", \n  Prefix: \"file-\"\n };\n s3.listObjectsV2(params, function(err, data) {\n   if (err) console.log(err, err.stack); // an error occurred\n   else {\n       query = \"Contents[?LastModified>'2019-02-06T05:34:12.000Z']\"\n       var results = jmespath.search(data,query);\n   }\n };\n\n"]], "1_corpus": "stackexchange"}
+{"tstamp": 1722266165.7978, "task_type": "retrieval", "type": "rightvote", "models": ["", ""], "ip": "", "0_conv_id": "cacdd0fd6204439b8b8081984a882bac", "0_model_name": "sentence-transformers/all-MiniLM-L6-v2", "0_prompt": "can myiasis kill a dog?", "0_output": [["can myiasis kill a dog?", "Stackoverflow Stackexchange\n\nQ: Visual C# backtrace: how to know where [external code] resides? I am using visual c# to debug a large c# project. The project is very large and only some of  the pdbs are build by myself. I can build all the pdb however it's very time consuming. I am debugging a backtrace but a part of the trace is marked with [External Code]. I know some pdb is missing, but how can I know which DLL the external code resides? In C I can get the EIP then see the process mapping to determine where the EIP lives, but how this can be done in .Net environment?\n\nA: Go into Tools | Options | Debugging | General, and uncheck the \"Enable Just My Code\" checkbox.  Now Visual Studio will show the full call stack (at least for managed code, you still may see some \"native to managed transitions\" in there).\n"]], "0_corpus": "stackexchange", "1_conv_id": "fed637e5f4d04a5fa833747983dfde2a", "1_model_name": "mixedbread-ai/mxbai-embed-large-v1", "1_prompt": "can myiasis kill a dog?", "1_output": [["can myiasis kill a dog?", "Gaming Stackexchange\n\nQ: Can My Pet Be Killed? Maybe I just haven't noticed, but it my pet never seems to take damage.  Can he be killed?  What happens if he can, and does die?\n\nA: No, your pet cannot die.  If your pet gets too low on life, they'll panic, run away and return when they recover.  They'll regenerate life slowly while fleeing, so if you wait a bit, they'll come back on their own.\nYou can speed the healing process by giving your pet a health potion.  I believe this requires that you shift-click or shift-hotkey (ie, if your health potion is 1, press Shift-1) a health potion.\nYou can also make your pet tougher by feeding it fish, so they don't take as much damage, and therefore don't run away as easily.  Different fish have different attributes and last for different amount of time, so experiment to find the one that works best for you, and then feed it that fish before/during a fight with a large boss or other dangerous area of the game.\n"]], "1_corpus": "stackexchange"}
+{"tstamp": 1722266265.9303, "task_type": "retrieval", "type": "rightvote", "models": ["", ""], "ip": "", "0_conv_id": "2c0c8080a45d4806a27966e78ac25d0a", "0_model_name": "voyage-multilingual-2", "0_prompt": "how long is one full rotation of the earth", "0_output": [["how long is one full rotation of the earth", "Earth's rotation\n\nThe apparent solar time is a measure of Earth's rotation and the difference between it and the mean solar time is known as the equation of time.\nStellar and sidereal day\nEarth's rotation period relative to the International Celestial Reference Frame, called its stellar day by the International Earth Rotation and Reference Systems Service (IERS), is  seconds of mean solar time (UT1) , ). Earth's rotation period relative to the precessing mean vernal equinox, named sidereal day, is  of mean solar time (UT1) , ). Thus, the sidereal day is shorter than the stellar day by about .\nBoth the stellar day and the sidereal day are shorter than the mean solar day by about  . This is a result of the Earth turning 1 additional rotation, relative to the celestial reference frame, as it orbits the Sun (so 366.24 rotations/y). The mean solar day in SI seconds is available from the IERS for the periods  and .\nRecently (1999\u20132010) the average annual length of the mean solar day in excess of 86,400 SI seconds has varied between  and , which must be added to both the stellar and sidereal days given in mean solar time above to obtain their lengths in SI seconds (see Fluctuations in the length of day)."]], "0_corpus": "wikipedia", "1_conv_id": "4a656424412845e18fdce95bcd7c5bc4", "1_model_name": "intfloat/multilingual-e5-large-instruct", "1_prompt": "how long is one full rotation of the earth", "1_output": [["how long is one full rotation of the earth", "Earth's rotation\n\nEarth rotates once in about 24 hours with respect to the Sun, but once every 23 hours, 56 minutes and 4 seconds with respect to other distant stars (see below). Earth's rotation is slowing slightly with time; thus, a day was shorter in the past. This is due to the tidal effects the Moon has on Earth's rotation. Atomic clocks show that the modern day is longer by about 1.7 milliseconds than a century ago, slowly increasing the rate at which UTC is adjusted by leap seconds. Analysis of historical astronomical records shows a slowing trend; the length of a day increased by about 2.3 milliseconds per century since the 8th century BCE.\nScientists reported that in 2020 Earth had started spinning faster, after consistently spinning slower than 86,400 seconds per day in the decades before. On June 29, 2022, Earth's spin was completed in 1.59 milliseconds under 24 hours, setting a new record. Because of that trend, engineers worldwide are discussing a 'negative leap second' and other possible timekeeping measures.\nThis increase in speed is thought to be due to various factors, including the complex motion of its molten core, oceans, and atmosphere, the effect of celestial bodies such as the Moon, and possibly climate change, which is causing the ice at Earth's poles to melt. The masses of ice account for the Earth's shape being that of an oblate spheroid, bulging around the equator. When these masses are reduced, the poles rebound from the loss of weight, and Earth becomes more spherical, which has the effect of bringing mass closer to its centre of gravity. Conservation of angular momentum dictates that a mass distributed more closely around its centre of gravity spins faster."]], "1_corpus": "wikipedia"}

data/retrieval_individual-1fa504af-967a-4ff8-a6ff-4abf5fbdd1c5.jsonl ADDED Viewed

	@@ -0,0 +1,6 @@

+{"tstamp": 1722266110.9664, "task_type": "retrieval", "type": "chat", "model": "mixedbread-ai/mxbai-embed-large-v1", "gen_params": {}, "start": 1722266094.4704, "finish": 1722266110.9664, "ip": "", "conv_id": "8c04b396b11b41daa58f4645afb37357", "model_name": "mixedbread-ai/mxbai-embed-large-v1", "prompt": "do dusk balls work in max raid dens?", "output": [["do dusk balls work in max raid dens?", "Gaming Stackexchange\n\nQ: Do turns carry over from previous encounters with fleeing Legendary Pok\u00e9mon? I've been trying to catch Thundurus, but I don't know which is the best Pok\u00e9 Ball, the Quick ball or the Ultra ball? \nI left the wild Thundurus with 1 hp and asleep, but I don't know if it counts as one turn on the new encounters or if my previous turns carry over. I am using both types of Pok\u00e9 Balls at the moment, and it does not look like any are good. >.<\n\nA: Despite the fact that you may have been pecking away at wandering pokemon for many, many turns, whenever you meet them, it starts a new battle, from turn 1.\nThis means that the Quick Ball has twice the capture rate (x4) than that of the Ultra Ball (x2).\n"]], "corpus": "stackexchange"}
+{"tstamp": 1722266110.9664, "task_type": "retrieval", "type": "chat", "model": "sentence-transformers/all-MiniLM-L6-v2", "gen_params": {}, "start": 1722266094.4704, "finish": 1722266110.9664, "ip": "", "conv_id": "ecb249a3187244458381cbb0af7b2bb2", "model_name": "sentence-transformers/all-MiniLM-L6-v2", "prompt": "do dusk balls work in max raid dens?", "output": [["do dusk balls work in max raid dens?", "Stackoverflow Stackexchange\n\nQ: AWS-SDK: Query parameter in listobjects of S3 I want to fetch objects after a particular date. Using the AWS CLI I can list objects using below command: \naws s3api list-objects-v2 --bucket \"bucket1\" --prefix \"file-\" --query \"(Contents[?LastModified>'2019-02-06T05:34:12.000Z'])[0]\"\nBut I want to do it from my code so please let me know how can I filter objects usin NPM AWS-SDK.\nNote: I can do it using exec or spawn but for this I have to configure profile using CLI which will create credential file on local so I dont want to do this.\n\nA: Use the AWS SDK for node.js. Call the listObjectsV2 method and then use jmespath.js in the callback method to filter the output of the API call. This is the equivalent of what the AWS CLI does via the --query param.\nSomething like this (untested)\nvar params = {\n  Bucket: \"bucket1\", \n  Prefix: \"file-\"\n };\n s3.listObjectsV2(params, function(err, data) {\n   if (err) console.log(err, err.stack); // an error occurred\n   else {\n       query = \"Contents[?LastModified>'2019-02-06T05:34:12.000Z']\"\n       var results = jmespath.search(data,query);\n   }\n };\n\n"]], "corpus": "stackexchange"}
+{"tstamp": 1722266153.5527, "task_type": "retrieval", "type": "chat", "model": "sentence-transformers/all-MiniLM-L6-v2", "gen_params": {}, "start": 1722266153.4268, "finish": 1722266153.5527, "ip": "", "conv_id": "cacdd0fd6204439b8b8081984a882bac", "model_name": "sentence-transformers/all-MiniLM-L6-v2", "prompt": "can myiasis kill a dog?", "output": [["can myiasis kill a dog?", "Stackoverflow Stackexchange\n\nQ: Visual C# backtrace: how to know where [external code] resides? I am using visual c# to debug a large c# project. The project is very large and only some of  the pdbs are build by myself. I can build all the pdb however it's very time consuming. I am debugging a backtrace but a part of the trace is marked with [External Code]. I know some pdb is missing, but how can I know which DLL the external code resides? In C I can get the EIP then see the process mapping to determine where the EIP lives, but how this can be done in .Net environment?\n\nA: Go into Tools | Options | Debugging | General, and uncheck the \"Enable Just My Code\" checkbox.  Now Visual Studio will show the full call stack (at least for managed code, you still may see some \"native to managed transitions\" in there).\n"]], "corpus": "stackexchange"}
+{"tstamp": 1722266153.5527, "task_type": "retrieval", "type": "chat", "model": "mixedbread-ai/mxbai-embed-large-v1", "gen_params": {}, "start": 1722266153.4268, "finish": 1722266153.5527, "ip": "", "conv_id": "fed637e5f4d04a5fa833747983dfde2a", "model_name": "mixedbread-ai/mxbai-embed-large-v1", "prompt": "can myiasis kill a dog?", "output": [["can myiasis kill a dog?", "Gaming Stackexchange\n\nQ: Can My Pet Be Killed? Maybe I just haven't noticed, but it my pet never seems to take damage.  Can he be killed?  What happens if he can, and does die?\n\nA: No, your pet cannot die.  If your pet gets too low on life, they'll panic, run away and return when they recover.  They'll regenerate life slowly while fleeing, so if you wait a bit, they'll come back on their own.\nYou can speed the healing process by giving your pet a health potion.  I believe this requires that you shift-click or shift-hotkey (ie, if your health potion is 1, press Shift-1) a health potion.\nYou can also make your pet tougher by feeding it fish, so they don't take as much damage, and therefore don't run away as easily.  Different fish have different attributes and last for different amount of time, so experiment to find the one that works best for you, and then feed it that fish before/during a fight with a large boss or other dangerous area of the game.\n"]], "corpus": "stackexchange"}
+{"tstamp": 1722266208.7762, "task_type": "retrieval", "type": "chat", "model": "voyage-multilingual-2", "gen_params": {}, "start": 1722266172.5065, "finish": 1722266208.7762, "ip": "", "conv_id": "2c0c8080a45d4806a27966e78ac25d0a", "model_name": "voyage-multilingual-2", "prompt": "how long is one full rotation of the earth", "output": [["how long is one full rotation of the earth", "Earth's rotation\n\nThe apparent solar time is a measure of Earth's rotation and the difference between it and the mean solar time is known as the equation of time.\nStellar and sidereal day\nEarth's rotation period relative to the International Celestial Reference Frame, called its stellar day by the International Earth Rotation and Reference Systems Service (IERS), is  seconds of mean solar time (UT1) , ). Earth's rotation period relative to the precessing mean vernal equinox, named sidereal day, is  of mean solar time (UT1) , ). Thus, the sidereal day is shorter than the stellar day by about .\nBoth the stellar day and the sidereal day are shorter than the mean solar day by about  . This is a result of the Earth turning 1 additional rotation, relative to the celestial reference frame, as it orbits the Sun (so 366.24 rotations/y). The mean solar day in SI seconds is available from the IERS for the periods  and .\nRecently (1999\u20132010) the average annual length of the mean solar day in excess of 86,400 SI seconds has varied between  and , which must be added to both the stellar and sidereal days given in mean solar time above to obtain their lengths in SI seconds (see Fluctuations in the length of day)."]], "corpus": "wikipedia"}
+{"tstamp": 1722266208.7762, "task_type": "retrieval", "type": "chat", "model": "intfloat/multilingual-e5-large-instruct", "gen_params": {}, "start": 1722266172.5065, "finish": 1722266208.7762, "ip": "", "conv_id": "4a656424412845e18fdce95bcd7c5bc4", "model_name": "intfloat/multilingual-e5-large-instruct", "prompt": "how long is one full rotation of the earth", "output": [["how long is one full rotation of the earth", "Earth's rotation\n\nEarth rotates once in about 24 hours with respect to the Sun, but once every 23 hours, 56 minutes and 4 seconds with respect to other distant stars (see below). Earth's rotation is slowing slightly with time; thus, a day was shorter in the past. This is due to the tidal effects the Moon has on Earth's rotation. Atomic clocks show that the modern day is longer by about 1.7 milliseconds than a century ago, slowly increasing the rate at which UTC is adjusted by leap seconds. Analysis of historical astronomical records shows a slowing trend; the length of a day increased by about 2.3 milliseconds per century since the 8th century BCE.\nScientists reported that in 2020 Earth had started spinning faster, after consistently spinning slower than 86,400 seconds per day in the decades before. On June 29, 2022, Earth's spin was completed in 1.59 milliseconds under 24 hours, setting a new record. Because of that trend, engineers worldwide are discussing a 'negative leap second' and other possible timekeeping measures.\nThis increase in speed is thought to be due to various factors, including the complex motion of its molten core, oceans, and atmosphere, the effect of celestial bodies such as the Moon, and possibly climate change, which is causing the ice at Earth's poles to melt. The masses of ice account for the Earth's shape being that of an oblate spheroid, bulging around the equator. When these masses are reduced, the poles rebound from the loss of weight, and Earth becomes more spherical, which has the effect of bringing mass closer to its centre of gravity. Conservation of angular momentum dictates that a mass distributed more closely around its centre of gravity spins faster."]], "corpus": "wikipedia"}