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R: Generate tables based on Distinct Values
Good Morning, Community:
I have this table, which contains demand for specific items, on a given location and a given date:
<blockquote class="imgur-embed-pub" lang="en" data-id="a/RAoXIbW"><a href="//imgur.com/RAoXIbW"></a></blockquote><script async src="//s.imgur.com/min/embed.js" charset="utf-8"></script>Assuming the demand is already grouped (Sum) by Date, Product ID and Location ID, is there a way of generating several tables based distinct Product IDs and Location IDs, containing their respective demand in R?
This how it would look like for product "a" on location "3":
<blockquote class="imgur-embed-pub" lang="en" data-id="a/wv1hBxP"><a href="//imgur.com/wv1hBxP"></a></blockquote><script async src="//s.imgur.com/min/embed.js" charset="utf-8"></script>Thanks in advance.
r
asked Nov 21 '18 at 18:10
J-SawJ-Saw
102
add a comment |
Good Morning, Community:
I have this table, which contains demand for specific items, on a given location and a given date:
<blockquote class="imgur-embed-pub" lang="en" data-id="a/RAoXIbW"><a href="//imgur.com/RAoXIbW"></a></blockquote><script async src="//s.imgur.com/min/embed.js" charset="utf-8"></script>Assuming the demand is already grouped (Sum) by Date, Product ID and Location ID, is there a way of generating several tables based distinct Product IDs and Location IDs, containing their respective demand in R?
This how it would look like for product "a" on location "3":
<blockquote class="imgur-embed-pub" lang="en" data-id="a/wv1hBxP"><a href="//imgur.com/wv1hBxP"></a></blockquote><script async src="//s.imgur.com/min/embed.js" charset="utf-8"></script>Thanks in advance.
r
asked Nov 21 '18 at 18:10
J-SawJ-Saw
102
add a comment |
Good Morning, Community:
I have this table, which contains demand for specific items, on a given location and a given date:
<blockquote class="imgur-embed-pub" lang="en" data-id="a/RAoXIbW"><a href="//imgur.com/RAoXIbW"></a></blockquote><script async src="//s.imgur.com/min/embed.js" charset="utf-8"></script>Assuming the demand is already grouped (Sum) by Date, Product ID and Location ID, is there a way of generating several tables based distinct Product IDs and Location IDs, containing their respective demand in R?
This how it would look like for product "a" on location "3":
<blockquote class="imgur-embed-pub" lang="en" data-id="a/wv1hBxP"><a href="//imgur.com/wv1hBxP"></a></blockquote><script async src="//s.imgur.com/min/embed.js" charset="utf-8"></script>Thanks in advance.
r
asked Nov 21 '18 at 18:10
J-SawJ-Saw
102
Good Morning, Community:
I have this table, which contains demand for specific items, on a given location and a given date:
<blockquote class="imgur-embed-pub" lang="en" data-id="a/RAoXIbW"><a href="//imgur.com/RAoXIbW"></a></blockquote><script async src="//s.imgur.com/min/embed.js" charset="utf-8"></script>Assuming the demand is already grouped (Sum) by Date, Product ID and Location ID, is there a way of generating several tables based distinct Product IDs and Location IDs, containing their respective demand in R?
This how it would look like for product "a" on location "3":
<blockquote class="imgur-embed-pub" lang="en" data-id="a/wv1hBxP"><a href="//imgur.com/wv1hBxP"></a></blockquote><script async src="//s.imgur.com/min/embed.js" charset="utf-8"></script>Thanks in advance.
<blockquote class="imgur-embed-pub" lang="en" data-id="a/RAoXIbW"><a href="//imgur.com/RAoXIbW"></a></blockquote><script async src="//s.imgur.com/min/embed.js" charset="utf-8"></script><blockquote class="imgur-embed-pub" lang="en" data-id="a/RAoXIbW"><a href="//imgur.com/RAoXIbW"></a></blockquote><script async src="//s.imgur.com/min/embed.js" charset="utf-8"></script><blockquote class="imgur-embed-pub" lang="en" data-id="a/wv1hBxP"><a href="//imgur.com/wv1hBxP"></a></blockquote><script async src="//s.imgur.com/min/embed.js" charset="utf-8"></script><blockquote class="imgur-embed-pub" lang="en" data-id="a/wv1hBxP"><a href="//imgur.com/wv1hBxP"></a></blockquote><script async src="//s.imgur.com/min/embed.js" charset="utf-8"></script>r
r
asked Nov 21 '18 at 18:10
J-SawJ-Saw
102
asked Nov 21 '18 at 18:10
J-SawJ-Saw
102
asked Nov 21 '18 at 18:10
J-SawJ-Saw
102
asked Nov 21 '18 at 18:10
J-SawJ-Saw
102
asked Nov 21 '18 at 18:10
J-SawJ-Saw
102
102
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
We can use split to split the datasets into several in a list and then we can extract different datasets from the list
lst1 <- split(df1, list(df1$ProductID, df1$LocationID), drop = TRUE)
answered Nov 21 '18 at 18:15
akrunakrun
408k13198273
1
Worked perfectly! Thank you so much.
– J-Saw
Nov 23 '18 at 3:13
add a comment |
Good evening,
if your data is stored in a dataframe, say df, then
df[df["Product ID"]=="a",]
returns the data filtered for product "a".
answered Nov 21 '18 at 18:16
otwtmotwtm
36110
Don't think so. Where?
– otwtm
Nov 21 '18 at 18:31
oh yes, thanks.
– otwtm
Nov 21 '18 at 18:33
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
We can use split to split the datasets into several in a list and then we can extract different datasets from the list
lst1 <- split(df1, list(df1$ProductID, df1$LocationID), drop = TRUE)
answered Nov 21 '18 at 18:15
akrunakrun
408k13198273
1
Worked perfectly! Thank you so much.
– J-Saw
Nov 23 '18 at 3:13
add a comment |
We can use split to split the datasets into several in a list and then we can extract different datasets from the list
lst1 <- split(df1, list(df1$ProductID, df1$LocationID), drop = TRUE)
answered Nov 21 '18 at 18:15
akrunakrun
408k13198273
1
Worked perfectly! Thank you so much.
– J-Saw
Nov 23 '18 at 3:13
add a comment |
We can use split to split the datasets into several in a list and then we can extract different datasets from the list
lst1 <- split(df1, list(df1$ProductID, df1$LocationID), drop = TRUE)
answered Nov 21 '18 at 18:15
akrunakrun
408k13198273
We can use split to split the datasets into several in a list and then we can extract different datasets from the list
lst1 <- split(df1, list(df1$ProductID, df1$LocationID), drop = TRUE)
answered Nov 21 '18 at 18:15
akrunakrun
408k13198273
answered Nov 21 '18 at 18:15
akrunakrun
408k13198273
answered Nov 21 '18 at 18:15
akrunakrun
408k13198273
answered Nov 21 '18 at 18:15
akrunakrun
408k13198273
408k13198273
1
Worked perfectly! Thank you so much.
– J-Saw
Nov 23 '18 at 3:13
add a comment |
1
Worked perfectly! Thank you so much.
– J-Saw
Nov 23 '18 at 3:13
1
1
Worked perfectly! Thank you so much.
– J-Saw
Nov 23 '18 at 3:13
Worked perfectly! Thank you so much.
– J-Saw
Nov 23 '18 at 3:13
add a comment |
Good evening,
if your data is stored in a dataframe, say df, then
df[df["Product ID"]=="a",]
returns the data filtered for product "a".
answered Nov 21 '18 at 18:16
otwtmotwtm
36110
Don't think so. Where?
– otwtm
Nov 21 '18 at 18:31
oh yes, thanks.
– otwtm
Nov 21 '18 at 18:33
add a comment |
Good evening,
if your data is stored in a dataframe, say df, then
df[df["Product ID"]=="a",]
returns the data filtered for product "a".
answered Nov 21 '18 at 18:16
otwtmotwtm
36110
Don't think so. Where?
– otwtm
Nov 21 '18 at 18:31
oh yes, thanks.
– otwtm
Nov 21 '18 at 18:33
add a comment |
Good evening,
if your data is stored in a dataframe, say df, then
df[df["Product ID"]=="a",]
returns the data filtered for product "a".
answered Nov 21 '18 at 18:16
otwtmotwtm
36110
Good evening,
if your data is stored in a dataframe, say df, then
df[df["Product ID"]=="a",]
returns the data filtered for product "a".
answered Nov 21 '18 at 18:16
otwtmotwtm
36110
edited Nov 21 '18 at 18:33
answered Nov 21 '18 at 18:16
otwtmotwtm
36110
answered Nov 21 '18 at 18:16
otwtmotwtm
36110
answered Nov 21 '18 at 18:16
otwtmotwtm
36110
36110
Don't think so. Where?
– otwtm
Nov 21 '18 at 18:31
oh yes, thanks.
– otwtm
Nov 21 '18 at 18:33
add a comment |
Don't think so. Where?
– otwtm
Nov 21 '18 at 18:31
oh yes, thanks.
– otwtm
Nov 21 '18 at 18:33
Don't think so. Where?
– otwtm
Nov 21 '18 at 18:31
Don't think so. Where?
– otwtm
Nov 21 '18 at 18:31
oh yes, thanks.
– otwtm
Nov 21 '18 at 18:33
oh yes, thanks.
– otwtm
Nov 21 '18 at 18:33
add a comment |
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