How to remove adjacent date strings in a list?

up vote
3
down vote

favorite

Given a list as

list = "2018-9-1",1,2,3,"2018-9-2","2018-9-3",4,5,6,"2018-9-5",7,8,9,"2018-9-6","2018-9-7",10,11,12,"2018-9-8",13,14,15,"2018-9-9",16,17,18

list = "2018-9-1","a","b","c","2018-9-2","2018-9-3","d","e","f","2018-9-5","h","i","j","2018-9-6","2018-9-7","k","l","m","2018-9-8","n","o","p","2018-9-9","q","r","s"

how can adjacent date strings being pruned to get this result?

"2018-9-1",1,2,3,"2018-9-3",4,5,6,"2018-9-5",7,8,9,"2018-9-7",10,11,12,"2018-9-8",13,14,15,"2018-9-9",16,17,18

"2018-9-1","a","b","c","2018-9-3","d","e","f","2018-9-5","h","i","j","2018-9-7","k","l","m","2018-9-8","n","o","p","2018-9-9","q","r","s"

such that the function Partition could be applied properly.
Thanks!

edited Sep 22 at 5:35

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

2,314725

asked Sep 22 at 4:14

Jerry

35819

4

Something like: SequenceReplace[list, _String, s_String -> s]?
â€“Â Carl Woll
Sep 22 at 4:17

@Call Woll Thanks!
â€“Â Jerry
Sep 22 at 4:24

add a commentÂ |Â

up vote
3
down vote

favorite

Given a list as

list = "2018-9-1",1,2,3,"2018-9-2","2018-9-3",4,5,6,"2018-9-5",7,8,9,"2018-9-6","2018-9-7",10,11,12,"2018-9-8",13,14,15,"2018-9-9",16,17,18

list = "2018-9-1","a","b","c","2018-9-2","2018-9-3","d","e","f","2018-9-5","h","i","j","2018-9-6","2018-9-7","k","l","m","2018-9-8","n","o","p","2018-9-9","q","r","s"

how can adjacent date strings being pruned to get this result?

"2018-9-1",1,2,3,"2018-9-3",4,5,6,"2018-9-5",7,8,9,"2018-9-7",10,11,12,"2018-9-8",13,14,15,"2018-9-9",16,17,18

"2018-9-1","a","b","c","2018-9-3","d","e","f","2018-9-5","h","i","j","2018-9-7","k","l","m","2018-9-8","n","o","p","2018-9-9","q","r","s"

such that the function Partition could be applied properly.
Thanks!

edited Sep 22 at 5:35

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

2,314725

asked Sep 22 at 4:14

Jerry

35819

4

Something like: SequenceReplace[list, _String, s_String -> s]?
â€“Â Carl Woll
Sep 22 at 4:17

@Call Woll Thanks!
â€“Â Jerry
Sep 22 at 4:24

add a commentÂ |Â

up vote
3
down vote

favorite

Given a list as

list = "2018-9-1",1,2,3,"2018-9-2","2018-9-3",4,5,6,"2018-9-5",7,8,9,"2018-9-6","2018-9-7",10,11,12,"2018-9-8",13,14,15,"2018-9-9",16,17,18

list = "2018-9-1","a","b","c","2018-9-2","2018-9-3","d","e","f","2018-9-5","h","i","j","2018-9-6","2018-9-7","k","l","m","2018-9-8","n","o","p","2018-9-9","q","r","s"

how can adjacent date strings being pruned to get this result?

"2018-9-1",1,2,3,"2018-9-3",4,5,6,"2018-9-5",7,8,9,"2018-9-7",10,11,12,"2018-9-8",13,14,15,"2018-9-9",16,17,18

"2018-9-1","a","b","c","2018-9-3","d","e","f","2018-9-5","h","i","j","2018-9-7","k","l","m","2018-9-8","n","o","p","2018-9-9","q","r","s"

such that the function Partition could be applied properly.
Thanks!

edited Sep 22 at 5:35

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

2,314725

asked Sep 22 at 4:14

Jerry

35819

Given a list as

list = "2018-9-1",1,2,3,"2018-9-2","2018-9-3",4,5,6,"2018-9-5",7,8,9,"2018-9-6","2018-9-7",10,11,12,"2018-9-8",13,14,15,"2018-9-9",16,17,18

list = "2018-9-1","a","b","c","2018-9-2","2018-9-3","d","e","f","2018-9-5","h","i","j","2018-9-6","2018-9-7","k","l","m","2018-9-8","n","o","p","2018-9-9","q","r","s"

how can adjacent date strings being pruned to get this result?

"2018-9-1",1,2,3,"2018-9-3",4,5,6,"2018-9-5",7,8,9,"2018-9-7",10,11,12,"2018-9-8",13,14,15,"2018-9-9",16,17,18

"2018-9-1","a","b","c","2018-9-3","d","e","f","2018-9-5","h","i","j","2018-9-7","k","l","m","2018-9-8","n","o","p","2018-9-9","q","r","s"

such that the function Partition could be applied properly.
Thanks!

list-manipulation string-manipulation stringreplace

edited Sep 22 at 5:35

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

2,314725

asked Sep 22 at 4:14

Jerry

35819

edited Sep 22 at 5:35

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

2,314725

asked Sep 22 at 4:14

Jerry

35819

edited Sep 22 at 5:35

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

2,314725

edited Sep 22 at 5:35

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

2,314725

edited Sep 22 at 5:35

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

2,314725

asked Sep 22 at 4:14

Jerry

35819

asked Sep 22 at 4:14

Jerry

35819

asked Sep 22 at 4:14

Jerry

35819

4

Something like: SequenceReplace[list, _String, s_String -> s]?
â€“Â Carl Woll
Sep 22 at 4:17

@Call Woll Thanks!
â€“Â Jerry
Sep 22 at 4:24

add a commentÂ |Â

4

Something like: SequenceReplace[list, _String, s_String -> s]?
â€“Â Carl Woll
Sep 22 at 4:17

@Call Woll Thanks!
â€“Â Jerry
Sep 22 at 4:24

Something like: SequenceReplace[list, _String, s_String -> s]?
â€“Â Carl Woll
Sep 22 at 4:17

@Call Woll Thanks!
â€“Â Jerry
Sep 22 at 4:24

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
5
down vote

accepted

There are many ways to accomplish this. One way uses SequenceReplace. First define a dateStringQ predicate:

dateStringQ = StringMatchQ[DatePattern["Year","Month","Day"]];

Your examples:

list1 = 
 "2018-9-1",1,2,3,"2018-9-2","2018-9-3",4,5,6,"2018-9-5",7,8,9,
 "2018-9-6","2018-9-7",10,11,12,"2018-9-8",13,14,15,"2018-9-9",16,17,18
;

list2 = 
 "2018-9-1","a","b","c","2018-9-2","2018-9-3","d","e","f","2018-9-5",
 "h","i","j","2018-9-6","2018-9-7","k","l","m","2018-9-8","n","o","p",
 "2018-9-9","q","r","s"
;

Using SequenceReplace:

SequenceReplace[
 list1,
 _String?dateStringQ, s_String?dateStringQ :> s
]

SequenceReplace[
 list2,
 _String?dateStringQ, s_String?dateStringQ :> s
]

"2018-9-1", 1, 2, 3, "2018-9-3", 4, 5, 6, "2018-9-5", 7, 8, 9, "2018-9-7",
10, 11, 12, "2018-9-8", 13, 14, 15, "2018-9-9", 16, 17, 18

"2018-9-1", "a", "b", "c", "2018-9-3", "d", "e", "f", "2018-9-5", "h", "i",
"j", "2018-9-7", "k", "l", "m", "2018-9-8", "n", "o", "p", "2018-9-9", "q",
"r", "s"

answered Sep 22 at 5:19

Carl Woll

59.4k278154

why SequenceReplace[list2, _String?StringMatchQ[ DatePattern["Year", "Month", "Day"]], s_String?StringMatchQ[DatePattern["Year", "Month", "Day"]] :> s] doesnot work?
â€“Â Jerry
Sep 22 at 5:48

@Jerry The issue is precedence, _String?StringMatchQ[DataPattern["Year", "Month", "Day"]] is parsed as (_String?StringMatchQ)[DatePattern["Year", "Month", "Day"]] not _String?(StringMatchQ[DatePattern["Year", "Month", "Day"]]).
â€“Â Carl Woll
Sep 22 at 5:52

I see, Thanks very much!:)
â€“Â Jerry
Sep 22 at 6:00

add a commentÂ |Â

up vote
3
down vote

SequenceReplace is suitable, as @CarlWoll pointed out in the comment. But it needs further elaboration to work with both of your lists. Below is my version

stringPattern = RegularExpression["\d4-\d-\d"];
stringMatchQ = If[StringQ[#], StringMatchQ[#, stringPattern], False] &;
rule = a_?stringMatchQ, b_?stringMatchQ :> b;

SequenceReplace[list1, rule]
SequenceReplace[list2, rule]

"2018-9-1", 1, 2, 3, "2018-9-3", 4, 5, 6, "2018-9-5", 7, 8, 9, "2018-9-7", 10, 11, 12, "2018-9-8", 13, 14, 15, "2018-9-9", 16, 17, 18

"2018-9-1", "a", "b", "c", "2018-9-3", "d", "e", "f", "2018-9-5", "h", "i", "j", "2018-9-7", "k", "l", "m", "2018-9-8", "n", "o", "p", "2018-9-9", "q", "r", "s"

edited Sep 22 at 5:32

answered Sep 22 at 5:02

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

2,314725

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
5
down vote

accepted

There are many ways to accomplish this. One way uses SequenceReplace. First define a dateStringQ predicate:

dateStringQ = StringMatchQ[DatePattern["Year","Month","Day"]];

Your examples:

list1 = 
 "2018-9-1",1,2,3,"2018-9-2","2018-9-3",4,5,6,"2018-9-5",7,8,9,
 "2018-9-6","2018-9-7",10,11,12,"2018-9-8",13,14,15,"2018-9-9",16,17,18
;

list2 = 
 "2018-9-1","a","b","c","2018-9-2","2018-9-3","d","e","f","2018-9-5",
 "h","i","j","2018-9-6","2018-9-7","k","l","m","2018-9-8","n","o","p",
 "2018-9-9","q","r","s"
;

Using SequenceReplace:

SequenceReplace[
 list1,
 _String?dateStringQ, s_String?dateStringQ :> s
]

SequenceReplace[
 list2,
 _String?dateStringQ, s_String?dateStringQ :> s
]

"2018-9-1", 1, 2, 3, "2018-9-3", 4, 5, 6, "2018-9-5", 7, 8, 9, "2018-9-7",
10, 11, 12, "2018-9-8", 13, 14, 15, "2018-9-9", 16, 17, 18

"2018-9-1", "a", "b", "c", "2018-9-3", "d", "e", "f", "2018-9-5", "h", "i",
"j", "2018-9-7", "k", "l", "m", "2018-9-8", "n", "o", "p", "2018-9-9", "q",
"r", "s"

answered Sep 22 at 5:19

Carl Woll

59.4k278154

why SequenceReplace[list2, _String?StringMatchQ[ DatePattern["Year", "Month", "Day"]], s_String?StringMatchQ[DatePattern["Year", "Month", "Day"]] :> s] doesnot work?
â€“Â Jerry
Sep 22 at 5:48

@Jerry The issue is precedence, _String?StringMatchQ[DataPattern["Year", "Month", "Day"]] is parsed as (_String?StringMatchQ)[DatePattern["Year", "Month", "Day"]] not _String?(StringMatchQ[DatePattern["Year", "Month", "Day"]]).
â€“Â Carl Woll
Sep 22 at 5:52

I see, Thanks very much!:)
â€“Â Jerry
Sep 22 at 6:00

add a commentÂ |Â

up vote
5
down vote

accepted

There are many ways to accomplish this. One way uses SequenceReplace. First define a dateStringQ predicate:

dateStringQ = StringMatchQ[DatePattern["Year","Month","Day"]];

Your examples:

list1 = 
 "2018-9-1",1,2,3,"2018-9-2","2018-9-3",4,5,6,"2018-9-5",7,8,9,
 "2018-9-6","2018-9-7",10,11,12,"2018-9-8",13,14,15,"2018-9-9",16,17,18
;

list2 = 
 "2018-9-1","a","b","c","2018-9-2","2018-9-3","d","e","f","2018-9-5",
 "h","i","j","2018-9-6","2018-9-7","k","l","m","2018-9-8","n","o","p",
 "2018-9-9","q","r","s"
;

Using SequenceReplace:

SequenceReplace[
 list1,
 _String?dateStringQ, s_String?dateStringQ :> s
]

SequenceReplace[
 list2,
 _String?dateStringQ, s_String?dateStringQ :> s
]

"2018-9-1", 1, 2, 3, "2018-9-3", 4, 5, 6, "2018-9-5", 7, 8, 9, "2018-9-7",
10, 11, 12, "2018-9-8", 13, 14, 15, "2018-9-9", 16, 17, 18

"2018-9-1", "a", "b", "c", "2018-9-3", "d", "e", "f", "2018-9-5", "h", "i",
"j", "2018-9-7", "k", "l", "m", "2018-9-8", "n", "o", "p", "2018-9-9", "q",
"r", "s"

answered Sep 22 at 5:19

Carl Woll

59.4k278154

why SequenceReplace[list2, _String?StringMatchQ[ DatePattern["Year", "Month", "Day"]], s_String?StringMatchQ[DatePattern["Year", "Month", "Day"]] :> s] doesnot work?
â€“Â Jerry
Sep 22 at 5:48

@Jerry The issue is precedence, _String?StringMatchQ[DataPattern["Year", "Month", "Day"]] is parsed as (_String?StringMatchQ)[DatePattern["Year", "Month", "Day"]] not _String?(StringMatchQ[DatePattern["Year", "Month", "Day"]]).
â€“Â Carl Woll
Sep 22 at 5:52

I see, Thanks very much!:)
â€“Â Jerry
Sep 22 at 6:00

add a commentÂ |Â

up vote
5
down vote

accepted

There are many ways to accomplish this. One way uses SequenceReplace. First define a dateStringQ predicate:

dateStringQ = StringMatchQ[DatePattern["Year","Month","Day"]];

Your examples:

list1 = 
 "2018-9-1",1,2,3,"2018-9-2","2018-9-3",4,5,6,"2018-9-5",7,8,9,
 "2018-9-6","2018-9-7",10,11,12,"2018-9-8",13,14,15,"2018-9-9",16,17,18
;

list2 = 
 "2018-9-1","a","b","c","2018-9-2","2018-9-3","d","e","f","2018-9-5",
 "h","i","j","2018-9-6","2018-9-7","k","l","m","2018-9-8","n","o","p",
 "2018-9-9","q","r","s"
;

Using SequenceReplace:

SequenceReplace[
 list1,
 _String?dateStringQ, s_String?dateStringQ :> s
]

SequenceReplace[
 list2,
 _String?dateStringQ, s_String?dateStringQ :> s
]

"2018-9-1", 1, 2, 3, "2018-9-3", 4, 5, 6, "2018-9-5", 7, 8, 9, "2018-9-7",
10, 11, 12, "2018-9-8", 13, 14, 15, "2018-9-9", 16, 17, 18

"2018-9-1", "a", "b", "c", "2018-9-3", "d", "e", "f", "2018-9-5", "h", "i",
"j", "2018-9-7", "k", "l", "m", "2018-9-8", "n", "o", "p", "2018-9-9", "q",
"r", "s"

answered Sep 22 at 5:19

Carl Woll

59.4k278154

There are many ways to accomplish this. One way uses SequenceReplace. First define a dateStringQ predicate:

dateStringQ = StringMatchQ[DatePattern["Year","Month","Day"]];

Your examples:

list1 = 
 "2018-9-1",1,2,3,"2018-9-2","2018-9-3",4,5,6,"2018-9-5",7,8,9,
 "2018-9-6","2018-9-7",10,11,12,"2018-9-8",13,14,15,"2018-9-9",16,17,18
;

list2 = 
 "2018-9-1","a","b","c","2018-9-2","2018-9-3","d","e","f","2018-9-5",
 "h","i","j","2018-9-6","2018-9-7","k","l","m","2018-9-8","n","o","p",
 "2018-9-9","q","r","s"
;

Using SequenceReplace:

SequenceReplace[
 list1,
 _String?dateStringQ, s_String?dateStringQ :> s
]

SequenceReplace[
 list2,
 _String?dateStringQ, s_String?dateStringQ :> s
]

"2018-9-1", 1, 2, 3, "2018-9-3", 4, 5, 6, "2018-9-5", 7, 8, 9, "2018-9-7",
10, 11, 12, "2018-9-8", 13, 14, 15, "2018-9-9", 16, 17, 18

"2018-9-1", "a", "b", "c", "2018-9-3", "d", "e", "f", "2018-9-5", "h", "i",
"j", "2018-9-7", "k", "l", "m", "2018-9-8", "n", "o", "p", "2018-9-9", "q",
"r", "s"

answered Sep 22 at 5:19

Carl Woll

59.4k278154

answered Sep 22 at 5:19

Carl Woll

59.4k278154

answered Sep 22 at 5:19

Carl Woll

59.4k278154

answered Sep 22 at 5:19

Carl Woll

59.4k278154

why SequenceReplace[list2, _String?StringMatchQ[ DatePattern["Year", "Month", "Day"]], s_String?StringMatchQ[DatePattern["Year", "Month", "Day"]] :> s] doesnot work?
â€“Â Jerry
Sep 22 at 5:48

@Jerry The issue is precedence, _String?StringMatchQ[DataPattern["Year", "Month", "Day"]] is parsed as (_String?StringMatchQ)[DatePattern["Year", "Month", "Day"]] not _String?(StringMatchQ[DatePattern["Year", "Month", "Day"]]).
â€“Â Carl Woll
Sep 22 at 5:52

I see, Thanks very much!:)
â€“Â Jerry
Sep 22 at 6:00

add a commentÂ |Â

why SequenceReplace[list2, _String?StringMatchQ[ DatePattern["Year", "Month", "Day"]], s_String?StringMatchQ[DatePattern["Year", "Month", "Day"]] :> s] doesnot work?
â€“Â Jerry
Sep 22 at 5:48

@Jerry The issue is precedence, _String?StringMatchQ[DataPattern["Year", "Month", "Day"]] is parsed as (_String?StringMatchQ)[DatePattern["Year", "Month", "Day"]] not _String?(StringMatchQ[DatePattern["Year", "Month", "Day"]]).
â€“Â Carl Woll
Sep 22 at 5:52

I see, Thanks very much!:)
â€“Â Jerry
Sep 22 at 6:00

why

SequenceReplace[list2, _String?StringMatchQ[ DatePattern["Year", "Month", "Day"]], s_String?StringMatchQ[DatePattern["Year", "Month", "Day"]] :> s]

doesnot work?
â€“Â Jerry
Sep 22 at 5:48

why

SequenceReplace[list2, _String?StringMatchQ[ DatePattern["Year", "Month", "Day"]], s_String?StringMatchQ[DatePattern["Year", "Month", "Day"]] :> s]

doesnot work?
â€“Â Jerry
Sep 22 at 5:48

@Jerry The issue is precedence, _String?StringMatchQ[DataPattern["Year", "Month", "Day"]] is parsed as (_String?StringMatchQ)[DatePattern["Year", "Month", "Day"]] not _String?(StringMatchQ[DatePattern["Year", "Month", "Day"]]).
â€“Â Carl Woll
Sep 22 at 5:52

I see, Thanks very much!:)
â€“Â Jerry
Sep 22 at 6:00

add a commentÂ |Â

up vote
3
down vote

SequenceReplace is suitable, as @CarlWoll pointed out in the comment. But it needs further elaboration to work with both of your lists. Below is my version

stringPattern = RegularExpression["\d4-\d-\d"];
stringMatchQ = If[StringQ[#], StringMatchQ[#, stringPattern], False] &;
rule = a_?stringMatchQ, b_?stringMatchQ :> b;

SequenceReplace[list1, rule]
SequenceReplace[list2, rule]

"2018-9-1", 1, 2, 3, "2018-9-3", 4, 5, 6, "2018-9-5", 7, 8, 9, "2018-9-7", 10, 11, 12, "2018-9-8", 13, 14, 15, "2018-9-9", 16, 17, 18

"2018-9-1", "a", "b", "c", "2018-9-3", "d", "e", "f", "2018-9-5", "h", "i", "j", "2018-9-7", "k", "l", "m", "2018-9-8", "n", "o", "p", "2018-9-9", "q", "r", "s"

edited Sep 22 at 5:32

answered Sep 22 at 5:02

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

2,314725

add a commentÂ |Â

up vote
3
down vote

SequenceReplace is suitable, as @CarlWoll pointed out in the comment. But it needs further elaboration to work with both of your lists. Below is my version

stringPattern = RegularExpression["\d4-\d-\d"];
stringMatchQ = If[StringQ[#], StringMatchQ[#, stringPattern], False] &;
rule = a_?stringMatchQ, b_?stringMatchQ :> b;

SequenceReplace[list1, rule]
SequenceReplace[list2, rule]

"2018-9-1", 1, 2, 3, "2018-9-3", 4, 5, 6, "2018-9-5", 7, 8, 9, "2018-9-7", 10, 11, 12, "2018-9-8", 13, 14, 15, "2018-9-9", 16, 17, 18

"2018-9-1", "a", "b", "c", "2018-9-3", "d", "e", "f", "2018-9-5", "h", "i", "j", "2018-9-7", "k", "l", "m", "2018-9-8", "n", "o", "p", "2018-9-9", "q", "r", "s"

edited Sep 22 at 5:32

answered Sep 22 at 5:02

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

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3
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SequenceReplace is suitable, as @CarlWoll pointed out in the comment. But it needs further elaboration to work with both of your lists. Below is my version

stringPattern = RegularExpression["\d4-\d-\d"];
stringMatchQ = If[StringQ[#], StringMatchQ[#, stringPattern], False] &;
rule = a_?stringMatchQ, b_?stringMatchQ :> b;

SequenceReplace[list1, rule]
SequenceReplace[list2, rule]

"2018-9-1", 1, 2, 3, "2018-9-3", 4, 5, 6, "2018-9-5", 7, 8, 9, "2018-9-7", 10, 11, 12, "2018-9-8", 13, 14, 15, "2018-9-9", 16, 17, 18

"2018-9-1", "a", "b", "c", "2018-9-3", "d", "e", "f", "2018-9-5", "h", "i", "j", "2018-9-7", "k", "l", "m", "2018-9-8", "n", "o", "p", "2018-9-9", "q", "r", "s"

edited Sep 22 at 5:32

answered Sep 22 at 5:02

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

2,314725

SequenceReplace is suitable, as @CarlWoll pointed out in the comment. But it needs further elaboration to work with both of your lists. Below is my version

stringPattern = RegularExpression["\d4-\d-\d"];
stringMatchQ = If[StringQ[#], StringMatchQ[#, stringPattern], False] &;
rule = a_?stringMatchQ, b_?stringMatchQ :> b;

SequenceReplace[list1, rule]
SequenceReplace[list2, rule]

"2018-9-1", 1, 2, 3, "2018-9-3", 4, 5, 6, "2018-9-5", 7, 8, 9, "2018-9-7", 10, 11, 12, "2018-9-8", 13, 14, 15, "2018-9-9", 16, 17, 18

"2018-9-1", "a", "b", "c", "2018-9-3", "d", "e", "f", "2018-9-5", "h", "i", "j", "2018-9-7", "k", "l", "m", "2018-9-8", "n", "o", "p", "2018-9-9", "q", "r", "s"

edited Sep 22 at 5:32

answered Sep 22 at 5:02

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

2,314725

edited Sep 22 at 5:32

answered Sep 22 at 5:02

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

2,314725

answered Sep 22 at 5:02

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

2,314725

answered Sep 22 at 5:02

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

2,314725

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