Snakemake Using expand with dictionary
I am writing this rule:
rule process_files:
input:
dataout=expand("dataset/sample.ref.state.case.myresult.name.tsv", name=my_list[wildcards.ref])
output:
"dataset/sample.ref.state.case.endresult.tsv"
shell:
do something ...
Were expand
will get value from dictionary my_dictionary
based on the ref
value. I used wildcards
like this my_dictionary[wildcards.ref]
. But it ends up with this error name 'wildcards' is not defined
my_dictionary
something like:A:[1,2,3], B:[s1,s2..].....
I could use
def myfun(wildcards):
return expand("dataset/sample.ref.state.case.myresult.name.tsv", name=my_dictionary[wildcards.ref])
and use myfun
as input , but this does not answer why I can not use expand in place directly
Any suggestion how to fix it?
expand snakemake
add a comment |
I am writing this rule:
rule process_files:
input:
dataout=expand("dataset/sample.ref.state.case.myresult.name.tsv", name=my_list[wildcards.ref])
output:
"dataset/sample.ref.state.case.endresult.tsv"
shell:
do something ...
Were expand
will get value from dictionary my_dictionary
based on the ref
value. I used wildcards
like this my_dictionary[wildcards.ref]
. But it ends up with this error name 'wildcards' is not defined
my_dictionary
something like:A:[1,2,3], B:[s1,s2..].....
I could use
def myfun(wildcards):
return expand("dataset/sample.ref.state.case.myresult.name.tsv", name=my_dictionary[wildcards.ref])
and use myfun
as input , but this does not answer why I can not use expand in place directly
Any suggestion how to fix it?
expand snakemake
add a comment |
I am writing this rule:
rule process_files:
input:
dataout=expand("dataset/sample.ref.state.case.myresult.name.tsv", name=my_list[wildcards.ref])
output:
"dataset/sample.ref.state.case.endresult.tsv"
shell:
do something ...
Were expand
will get value from dictionary my_dictionary
based on the ref
value. I used wildcards
like this my_dictionary[wildcards.ref]
. But it ends up with this error name 'wildcards' is not defined
my_dictionary
something like:A:[1,2,3], B:[s1,s2..].....
I could use
def myfun(wildcards):
return expand("dataset/sample.ref.state.case.myresult.name.tsv", name=my_dictionary[wildcards.ref])
and use myfun
as input , but this does not answer why I can not use expand in place directly
Any suggestion how to fix it?
expand snakemake
I am writing this rule:
rule process_files:
input:
dataout=expand("dataset/sample.ref.state.case.myresult.name.tsv", name=my_list[wildcards.ref])
output:
"dataset/sample.ref.state.case.endresult.tsv"
shell:
do something ...
Were expand
will get value from dictionary my_dictionary
based on the ref
value. I used wildcards
like this my_dictionary[wildcards.ref]
. But it ends up with this error name 'wildcards' is not defined
my_dictionary
something like:A:[1,2,3], B:[s1,s2..].....
I could use
def myfun(wildcards):
return expand("dataset/sample.ref.state.case.myresult.name.tsv", name=my_dictionary[wildcards.ref])
and use myfun
as input , but this does not answer why I can not use expand in place directly
Any suggestion how to fix it?
expand snakemake
expand snakemake
edited Nov 14 '18 at 15:22
Medhat
asked Nov 13 '18 at 22:26
MedhatMedhat
940820
940820
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Your question seems similar to snakemake wildcards or expand command and the bottom line is that wildcards
is not defined in the input. So your solution of using an input function (or a lambda function) seems correct.
(As to why wildcards
is not defined in input, I don't know...)
Thanks, The function worked but it did not resolve the variable between double curly braces so it will ask for input fordataset/sample.ref.state.case
and raise an error.
– Medhat
Nov 14 '18 at 16:37
add a comment |
As @dariober mentioned there is the wildcards
objects but this is only accesible in the run/shell portion but can be accessed using an input function in input
.
Here is an example implementation that will expand the input based on the wildcards.ref
:
rule all:
input: expand("dataset/sample.ref.state.case.endresult.tsv", dataset=["D1", "D2"], sample=["S1", "S2"], ref=["R1", "R2"], state=["STATE1", "STATE2"], case=["C1", "C2"])
my_list = "R1": [1, 2, 3], "R2": ["s1", "s2"]
rule process_files:
input:
lambda wildcards: expand(
"dataset/sample.ref.state.case.myresult.name.tsv", name=my_list[wildcards.ref])
output:
"dataset/sample.ref.state.case.endresult.tsv"
shell:
"echo 'input' > output"
If you implement it as the lambda
function example above, it should resolve the issue you mention:
The function worked but it did not resolve the variable between double curly braces so it will ask for input for dataset/sample.ref.state.caseand raise an error.
Actually my function is the same as your lambda function and raises this error. defmyfun(wildcards): return expand("dataset/sample.ref.state.case.myresult.name.tsv", name=my_list[wildcards.ref])
. to overcome the issue I need to resolve each var for exampleref
. would be `wildcards.ref`` and so on.
– Medhat
Nov 14 '18 at 19:38
There should not really be need to do that. You are saying you pass to expand, in the case ofdataset
,dataset = wildcards.dataset
? Seems redundant. I am using snakemake 5.3.0 in the example and it works using yourmyfun
or lambda.
– JohnnyBD
Nov 14 '18 at 20:43
The issue is after using expand; the variable passed tosample
issample
so it would besample=sample
not the actual value of sample, which makes problem in processing for next step because now there is nothing calleddataset/sample...
in the input file
– Medhat
Nov 14 '18 at 21:23
I am sorry but I cannot seem to reproduce this issue you are mentioning. Could you maybe edit your question and provide example of what an input would look like for one input wildcard combination? Either I am misunderstanding what are you trying to do or our implementations are different? You want to have a single value for all the wildcards exceptname
? Essentially group a set ofname
inputs together? In that case you should havesample
in the result of expand as that wildcard will be deduced fromrule all
and output.
– JohnnyBD
Nov 14 '18 at 22:26
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
Your question seems similar to snakemake wildcards or expand command and the bottom line is that wildcards
is not defined in the input. So your solution of using an input function (or a lambda function) seems correct.
(As to why wildcards
is not defined in input, I don't know...)
Thanks, The function worked but it did not resolve the variable between double curly braces so it will ask for input fordataset/sample.ref.state.case
and raise an error.
– Medhat
Nov 14 '18 at 16:37
add a comment |
Your question seems similar to snakemake wildcards or expand command and the bottom line is that wildcards
is not defined in the input. So your solution of using an input function (or a lambda function) seems correct.
(As to why wildcards
is not defined in input, I don't know...)
Thanks, The function worked but it did not resolve the variable between double curly braces so it will ask for input fordataset/sample.ref.state.case
and raise an error.
– Medhat
Nov 14 '18 at 16:37
add a comment |
Your question seems similar to snakemake wildcards or expand command and the bottom line is that wildcards
is not defined in the input. So your solution of using an input function (or a lambda function) seems correct.
(As to why wildcards
is not defined in input, I don't know...)
Your question seems similar to snakemake wildcards or expand command and the bottom line is that wildcards
is not defined in the input. So your solution of using an input function (or a lambda function) seems correct.
(As to why wildcards
is not defined in input, I don't know...)
answered Nov 14 '18 at 8:25
darioberdariober
1,0361221
1,0361221
Thanks, The function worked but it did not resolve the variable between double curly braces so it will ask for input fordataset/sample.ref.state.case
and raise an error.
– Medhat
Nov 14 '18 at 16:37
add a comment |
Thanks, The function worked but it did not resolve the variable between double curly braces so it will ask for input fordataset/sample.ref.state.case
and raise an error.
– Medhat
Nov 14 '18 at 16:37
Thanks, The function worked but it did not resolve the variable between double curly braces so it will ask for input for
dataset/sample.ref.state.case
and raise an error.– Medhat
Nov 14 '18 at 16:37
Thanks, The function worked but it did not resolve the variable between double curly braces so it will ask for input for
dataset/sample.ref.state.case
and raise an error.– Medhat
Nov 14 '18 at 16:37
add a comment |
As @dariober mentioned there is the wildcards
objects but this is only accesible in the run/shell portion but can be accessed using an input function in input
.
Here is an example implementation that will expand the input based on the wildcards.ref
:
rule all:
input: expand("dataset/sample.ref.state.case.endresult.tsv", dataset=["D1", "D2"], sample=["S1", "S2"], ref=["R1", "R2"], state=["STATE1", "STATE2"], case=["C1", "C2"])
my_list = "R1": [1, 2, 3], "R2": ["s1", "s2"]
rule process_files:
input:
lambda wildcards: expand(
"dataset/sample.ref.state.case.myresult.name.tsv", name=my_list[wildcards.ref])
output:
"dataset/sample.ref.state.case.endresult.tsv"
shell:
"echo 'input' > output"
If you implement it as the lambda
function example above, it should resolve the issue you mention:
The function worked but it did not resolve the variable between double curly braces so it will ask for input for dataset/sample.ref.state.caseand raise an error.
Actually my function is the same as your lambda function and raises this error. defmyfun(wildcards): return expand("dataset/sample.ref.state.case.myresult.name.tsv", name=my_list[wildcards.ref])
. to overcome the issue I need to resolve each var for exampleref
. would be `wildcards.ref`` and so on.
– Medhat
Nov 14 '18 at 19:38
There should not really be need to do that. You are saying you pass to expand, in the case ofdataset
,dataset = wildcards.dataset
? Seems redundant. I am using snakemake 5.3.0 in the example and it works using yourmyfun
or lambda.
– JohnnyBD
Nov 14 '18 at 20:43
The issue is after using expand; the variable passed tosample
issample
so it would besample=sample
not the actual value of sample, which makes problem in processing for next step because now there is nothing calleddataset/sample...
in the input file
– Medhat
Nov 14 '18 at 21:23
I am sorry but I cannot seem to reproduce this issue you are mentioning. Could you maybe edit your question and provide example of what an input would look like for one input wildcard combination? Either I am misunderstanding what are you trying to do or our implementations are different? You want to have a single value for all the wildcards exceptname
? Essentially group a set ofname
inputs together? In that case you should havesample
in the result of expand as that wildcard will be deduced fromrule all
and output.
– JohnnyBD
Nov 14 '18 at 22:26
add a comment |
As @dariober mentioned there is the wildcards
objects but this is only accesible in the run/shell portion but can be accessed using an input function in input
.
Here is an example implementation that will expand the input based on the wildcards.ref
:
rule all:
input: expand("dataset/sample.ref.state.case.endresult.tsv", dataset=["D1", "D2"], sample=["S1", "S2"], ref=["R1", "R2"], state=["STATE1", "STATE2"], case=["C1", "C2"])
my_list = "R1": [1, 2, 3], "R2": ["s1", "s2"]
rule process_files:
input:
lambda wildcards: expand(
"dataset/sample.ref.state.case.myresult.name.tsv", name=my_list[wildcards.ref])
output:
"dataset/sample.ref.state.case.endresult.tsv"
shell:
"echo 'input' > output"
If you implement it as the lambda
function example above, it should resolve the issue you mention:
The function worked but it did not resolve the variable between double curly braces so it will ask for input for dataset/sample.ref.state.caseand raise an error.
Actually my function is the same as your lambda function and raises this error. defmyfun(wildcards): return expand("dataset/sample.ref.state.case.myresult.name.tsv", name=my_list[wildcards.ref])
. to overcome the issue I need to resolve each var for exampleref
. would be `wildcards.ref`` and so on.
– Medhat
Nov 14 '18 at 19:38
There should not really be need to do that. You are saying you pass to expand, in the case ofdataset
,dataset = wildcards.dataset
? Seems redundant. I am using snakemake 5.3.0 in the example and it works using yourmyfun
or lambda.
– JohnnyBD
Nov 14 '18 at 20:43
The issue is after using expand; the variable passed tosample
issample
so it would besample=sample
not the actual value of sample, which makes problem in processing for next step because now there is nothing calleddataset/sample...
in the input file
– Medhat
Nov 14 '18 at 21:23
I am sorry but I cannot seem to reproduce this issue you are mentioning. Could you maybe edit your question and provide example of what an input would look like for one input wildcard combination? Either I am misunderstanding what are you trying to do or our implementations are different? You want to have a single value for all the wildcards exceptname
? Essentially group a set ofname
inputs together? In that case you should havesample
in the result of expand as that wildcard will be deduced fromrule all
and output.
– JohnnyBD
Nov 14 '18 at 22:26
add a comment |
As @dariober mentioned there is the wildcards
objects but this is only accesible in the run/shell portion but can be accessed using an input function in input
.
Here is an example implementation that will expand the input based on the wildcards.ref
:
rule all:
input: expand("dataset/sample.ref.state.case.endresult.tsv", dataset=["D1", "D2"], sample=["S1", "S2"], ref=["R1", "R2"], state=["STATE1", "STATE2"], case=["C1", "C2"])
my_list = "R1": [1, 2, 3], "R2": ["s1", "s2"]
rule process_files:
input:
lambda wildcards: expand(
"dataset/sample.ref.state.case.myresult.name.tsv", name=my_list[wildcards.ref])
output:
"dataset/sample.ref.state.case.endresult.tsv"
shell:
"echo 'input' > output"
If you implement it as the lambda
function example above, it should resolve the issue you mention:
The function worked but it did not resolve the variable between double curly braces so it will ask for input for dataset/sample.ref.state.caseand raise an error.
As @dariober mentioned there is the wildcards
objects but this is only accesible in the run/shell portion but can be accessed using an input function in input
.
Here is an example implementation that will expand the input based on the wildcards.ref
:
rule all:
input: expand("dataset/sample.ref.state.case.endresult.tsv", dataset=["D1", "D2"], sample=["S1", "S2"], ref=["R1", "R2"], state=["STATE1", "STATE2"], case=["C1", "C2"])
my_list = "R1": [1, 2, 3], "R2": ["s1", "s2"]
rule process_files:
input:
lambda wildcards: expand(
"dataset/sample.ref.state.case.myresult.name.tsv", name=my_list[wildcards.ref])
output:
"dataset/sample.ref.state.case.endresult.tsv"
shell:
"echo 'input' > output"
If you implement it as the lambda
function example above, it should resolve the issue you mention:
The function worked but it did not resolve the variable between double curly braces so it will ask for input for dataset/sample.ref.state.caseand raise an error.
answered Nov 14 '18 at 17:47
JohnnyBDJohnnyBD
11115
11115
Actually my function is the same as your lambda function and raises this error. defmyfun(wildcards): return expand("dataset/sample.ref.state.case.myresult.name.tsv", name=my_list[wildcards.ref])
. to overcome the issue I need to resolve each var for exampleref
. would be `wildcards.ref`` and so on.
– Medhat
Nov 14 '18 at 19:38
There should not really be need to do that. You are saying you pass to expand, in the case ofdataset
,dataset = wildcards.dataset
? Seems redundant. I am using snakemake 5.3.0 in the example and it works using yourmyfun
or lambda.
– JohnnyBD
Nov 14 '18 at 20:43
The issue is after using expand; the variable passed tosample
issample
so it would besample=sample
not the actual value of sample, which makes problem in processing for next step because now there is nothing calleddataset/sample...
in the input file
– Medhat
Nov 14 '18 at 21:23
I am sorry but I cannot seem to reproduce this issue you are mentioning. Could you maybe edit your question and provide example of what an input would look like for one input wildcard combination? Either I am misunderstanding what are you trying to do or our implementations are different? You want to have a single value for all the wildcards exceptname
? Essentially group a set ofname
inputs together? In that case you should havesample
in the result of expand as that wildcard will be deduced fromrule all
and output.
– JohnnyBD
Nov 14 '18 at 22:26
add a comment |
Actually my function is the same as your lambda function and raises this error. defmyfun(wildcards): return expand("dataset/sample.ref.state.case.myresult.name.tsv", name=my_list[wildcards.ref])
. to overcome the issue I need to resolve each var for exampleref
. would be `wildcards.ref`` and so on.
– Medhat
Nov 14 '18 at 19:38
There should not really be need to do that. You are saying you pass to expand, in the case ofdataset
,dataset = wildcards.dataset
? Seems redundant. I am using snakemake 5.3.0 in the example and it works using yourmyfun
or lambda.
– JohnnyBD
Nov 14 '18 at 20:43
The issue is after using expand; the variable passed tosample
issample
so it would besample=sample
not the actual value of sample, which makes problem in processing for next step because now there is nothing calleddataset/sample...
in the input file
– Medhat
Nov 14 '18 at 21:23
I am sorry but I cannot seem to reproduce this issue you are mentioning. Could you maybe edit your question and provide example of what an input would look like for one input wildcard combination? Either I am misunderstanding what are you trying to do or our implementations are different? You want to have a single value for all the wildcards exceptname
? Essentially group a set ofname
inputs together? In that case you should havesample
in the result of expand as that wildcard will be deduced fromrule all
and output.
– JohnnyBD
Nov 14 '18 at 22:26
Actually my function is the same as your lambda function and raises this error. def
myfun(wildcards): return expand("dataset/sample.ref.state.case.myresult.name.tsv", name=my_list[wildcards.ref])
. to overcome the issue I need to resolve each var for example ref
. would be `wildcards.ref`` and so on.– Medhat
Nov 14 '18 at 19:38
Actually my function is the same as your lambda function and raises this error. def
myfun(wildcards): return expand("dataset/sample.ref.state.case.myresult.name.tsv", name=my_list[wildcards.ref])
. to overcome the issue I need to resolve each var for example ref
. would be `wildcards.ref`` and so on.– Medhat
Nov 14 '18 at 19:38
There should not really be need to do that. You are saying you pass to expand, in the case of
dataset
, dataset = wildcards.dataset
? Seems redundant. I am using snakemake 5.3.0 in the example and it works using your myfun
or lambda.– JohnnyBD
Nov 14 '18 at 20:43
There should not really be need to do that. You are saying you pass to expand, in the case of
dataset
, dataset = wildcards.dataset
? Seems redundant. I am using snakemake 5.3.0 in the example and it works using your myfun
or lambda.– JohnnyBD
Nov 14 '18 at 20:43
The issue is after using expand; the variable passed to
sample
is sample
so it would be sample=sample
not the actual value of sample, which makes problem in processing for next step because now there is nothing called dataset/sample...
in the input file– Medhat
Nov 14 '18 at 21:23
The issue is after using expand; the variable passed to
sample
is sample
so it would be sample=sample
not the actual value of sample, which makes problem in processing for next step because now there is nothing called dataset/sample...
in the input file– Medhat
Nov 14 '18 at 21:23
I am sorry but I cannot seem to reproduce this issue you are mentioning. Could you maybe edit your question and provide example of what an input would look like for one input wildcard combination? Either I am misunderstanding what are you trying to do or our implementations are different? You want to have a single value for all the wildcards except
name
? Essentially group a set of name
inputs together? In that case you should have sample
in the result of expand as that wildcard will be deduced from rule all
and output.– JohnnyBD
Nov 14 '18 at 22:26
I am sorry but I cannot seem to reproduce this issue you are mentioning. Could you maybe edit your question and provide example of what an input would look like for one input wildcard combination? Either I am misunderstanding what are you trying to do or our implementations are different? You want to have a single value for all the wildcards except
name
? Essentially group a set of name
inputs together? In that case you should have sample
in the result of expand as that wildcard will be deduced from rule all
and output.– JohnnyBD
Nov 14 '18 at 22:26
add a comment |
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