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What does a/.b[c][[1]] mean?
What are the most common pitfalls awaiting new users?What Method options does FindInstance accept?Using Element[]How to sort the functions in a Package?What does IndexBy do and how exactly does it work?How to count the number of values in a discrete function?Dealing with a sequence of function arguments that may contain listsWhat does empty mean in Module?Getting a list of function definition parameter valuesWhen does Root have a third argumentHow can I make assignments persist across sessions?
$begingroup$
The function b[c_] has been previously defined. Does that "1" refer to the smallest number of a list?
functions
asked 9 hours ago
user65725user65725
212
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add a comment |
$begingroup$
The function b[c_] has been previously defined. Does that "1" refer to the smallest number of a list?
functions
asked 9 hours ago
user65725user65725
212
New contributor
user65725 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
See mathematica.stackexchange.com/a/25616/193
$endgroup$
– Michael E2
8 hours ago
$begingroup$
Actually, "1" refers to the first element of a list, not the smallest number of a list. Thusa[[1]]is the same asFirst[a].
$endgroup$
– Somos
3 hours ago
add a comment |
$begingroup$
The function b[c_] has been previously defined. Does that "1" refer to the smallest number of a list?
functions
asked 9 hours ago
user65725user65725
212
New contributor
user65725 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The function b[c_] has been previously defined. Does that "1" refer to the smallest number of a list?
functions
functions
asked 9 hours ago
user65725user65725
212
New contributor
user65725 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
user65725user65725
212
New contributor
user65725 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
user65725user65725
212
New contributor
user65725 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
user65725user65725
212
asked 9 hours ago
user65725user65725
212
212
New contributor
user65725 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user65725 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
See mathematica.stackexchange.com/a/25616/193
$endgroup$
– Michael E2
8 hours ago
$begingroup$
Actually, "1" refers to the first element of a list, not the smallest number of a list. Thusa[[1]]is the same asFirst[a].
$endgroup$
– Somos
3 hours ago
add a comment |
$begingroup$
See mathematica.stackexchange.com/a/25616/193
$endgroup$
– Michael E2
8 hours ago
$begingroup$
Actually, "1" refers to the first element of a list, not the smallest number of a list. Thusa[[1]]is the same asFirst[a].
$endgroup$
– Somos
3 hours ago
$begingroup$
See mathematica.stackexchange.com/a/25616/193
$endgroup$
– Michael E2
8 hours ago
$begingroup$
See mathematica.stackexchange.com/a/25616/193
$endgroup$
– Michael E2
8 hours ago
$begingroup$
Actually, "1" refers to the first element of a list, not the smallest number of a list. Thus
a[[1]] is the same as First[a].$endgroup$
– Somos
3 hours ago
$begingroup$
Actually, "1" refers to the first element of a list, not the smallest number of a list. Thus
a[[1]] is the same as First[a].$endgroup$
– Somos
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you just want to know what the double brackets mean, you can highlight them and use help. Or, you can use Information (?), for example:
?[[
Or, you can use Hold with FullForm to figure out what it means:
a /. b[c][[1]] //Hold //FullForm
Hold[ReplaceAll[a,Part[b[c],1]]]
answered 9 hours ago
Carl WollCarl Woll
83.9k3106217
$endgroup$
add a comment |
$begingroup$
Yes, 1 refers to the first element of the list. "/." is a rule replacement so I presume in this case that function "b" is returning a list of rules, which the line of code intends to use as a replacement operator for a.
Starting list or array notation at "1" is sometimes confusing for people used to "c" starting indices at 0. Note that there is a form of "Mod" that will properly provide the right index so you don't end up with "0" when you try to get a part of a list.
answered 8 hours ago
Mark RMark R
1937
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
If you just want to know what the double brackets mean, you can highlight them and use help. Or, you can use Information (?), for example:
?[[
Or, you can use Hold with FullForm to figure out what it means:
a /. b[c][[1]] //Hold //FullForm
Hold[ReplaceAll[a,Part[b[c],1]]]
answered 9 hours ago
Carl WollCarl Woll
83.9k3106217
$endgroup$
add a comment |
$begingroup$
If you just want to know what the double brackets mean, you can highlight them and use help. Or, you can use Information (?), for example:
?[[
Or, you can use Hold with FullForm to figure out what it means:
a /. b[c][[1]] //Hold //FullForm
Hold[ReplaceAll[a,Part[b[c],1]]]
answered 9 hours ago
Carl WollCarl Woll
83.9k3106217
$endgroup$
add a comment |
$begingroup$
If you just want to know what the double brackets mean, you can highlight them and use help. Or, you can use Information (?), for example:
?[[
Or, you can use Hold with FullForm to figure out what it means:
a /. b[c][[1]] //Hold //FullForm
Hold[ReplaceAll[a,Part[b[c],1]]]
answered 9 hours ago
Carl WollCarl Woll
83.9k3106217
$endgroup$
If you just want to know what the double brackets mean, you can highlight them and use help. Or, you can use Information (?), for example:
?[[
Or, you can use Hold with FullForm to figure out what it means:
a /. b[c][[1]] //Hold //FullForm
Hold[ReplaceAll[a,Part[b[c],1]]]
answered 9 hours ago
Carl WollCarl Woll
83.9k3106217
answered 9 hours ago
Carl WollCarl Woll
83.9k3106217
answered 9 hours ago
Carl WollCarl Woll
83.9k3106217
answered 9 hours ago
Carl WollCarl Woll
83.9k3106217
83.9k3106217
add a comment |
add a comment |
$begingroup$
Yes, 1 refers to the first element of the list. "/." is a rule replacement so I presume in this case that function "b" is returning a list of rules, which the line of code intends to use as a replacement operator for a.
Starting list or array notation at "1" is sometimes confusing for people used to "c" starting indices at 0. Note that there is a form of "Mod" that will properly provide the right index so you don't end up with "0" when you try to get a part of a list.
answered 8 hours ago
Mark RMark R
1937
$endgroup$
add a comment |
$begingroup$
Yes, 1 refers to the first element of the list. "/." is a rule replacement so I presume in this case that function "b" is returning a list of rules, which the line of code intends to use as a replacement operator for a.
Starting list or array notation at "1" is sometimes confusing for people used to "c" starting indices at 0. Note that there is a form of "Mod" that will properly provide the right index so you don't end up with "0" when you try to get a part of a list.
answered 8 hours ago
Mark RMark R
1937
$endgroup$
add a comment |
$begingroup$
Yes, 1 refers to the first element of the list. "/." is a rule replacement so I presume in this case that function "b" is returning a list of rules, which the line of code intends to use as a replacement operator for a.
Starting list or array notation at "1" is sometimes confusing for people used to "c" starting indices at 0. Note that there is a form of "Mod" that will properly provide the right index so you don't end up with "0" when you try to get a part of a list.
answered 8 hours ago
Mark RMark R
1937
$endgroup$
Yes, 1 refers to the first element of the list. "/." is a rule replacement so I presume in this case that function "b" is returning a list of rules, which the line of code intends to use as a replacement operator for a.
Starting list or array notation at "1" is sometimes confusing for people used to "c" starting indices at 0. Note that there is a form of "Mod" that will properly provide the right index so you don't end up with "0" when you try to get a part of a list.
answered 8 hours ago
Mark RMark R
1937
answered 8 hours ago
Mark RMark R
1937
answered 8 hours ago
Mark RMark R
1937
answered 8 hours ago
Mark RMark R
1937
1937
add a comment |
add a comment |
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$begingroup$
See mathematica.stackexchange.com/a/25616/193
$endgroup$
– Michael E2
8 hours ago
$begingroup$
Actually, "1" refers to the first element of a list, not the smallest number of a list. Thus
a[[1]]is the same asFirst[a].$endgroup$
– Somos
3 hours ago