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Replacing items in a list in a dictionary with items from another dictionary
How to merge two dictionaries in a single expression?How do I check if a list is empty?How do I sort a list of dictionaries by a value of the dictionary?What is the best way to iterate over a dictionary?Finding the index of an item given a list containing it in PythonHow do I sort a dictionary by value?Add new keys to a dictionary?Check if a given key already exists in a dictionaryHow do I list all files of a directory?Iterating over dictionaries using 'for' loops
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
I have the following dictionaries:
d1 = "00f_5" :[1,2,3], "00f_6": [1,2,3]
d2 = ["marker":"00f_5",1: 'AAA',"marker":"00f_6", 1: 'CCC',"marker":"00f_5", 2:"AAC"]
I would like the following output:
d1 = "00f_5" :["AAA","AAC",3], "00f_6": ["CCC",2,3]
I have tried multiple attempts, but I couldn't get it. Any help would be much appreciated.
python python-3.x dictionary
asked Mar 9 at 3:25
Mandie DriskillMandie Driskill
386
add a comment |
I have the following dictionaries:
d1 = "00f_5" :[1,2,3], "00f_6": [1,2,3]
d2 = ["marker":"00f_5",1: 'AAA',"marker":"00f_6", 1: 'CCC',"marker":"00f_5", 2:"AAC"]
I would like the following output:
d1 = "00f_5" :["AAA","AAC",3], "00f_6": ["CCC",2,3]
I have tried multiple attempts, but I couldn't get it. Any help would be much appreciated.
python python-3.x dictionary
asked Mar 9 at 3:25
Mandie DriskillMandie Driskill
386
1
Please post what you have attempted so far, and where did you encounter problems. [SO]: How to create a Minimal, Complete, and Verifiable example (mcve).
– CristiFati
Mar 9 at 3:29
add a comment |
I have the following dictionaries:
d1 = "00f_5" :[1,2,3], "00f_6": [1,2,3]
d2 = ["marker":"00f_5",1: 'AAA',"marker":"00f_6", 1: 'CCC',"marker":"00f_5", 2:"AAC"]
I would like the following output:
d1 = "00f_5" :["AAA","AAC",3], "00f_6": ["CCC",2,3]
I have tried multiple attempts, but I couldn't get it. Any help would be much appreciated.
python python-3.x dictionary
asked Mar 9 at 3:25
Mandie DriskillMandie Driskill
386
I have the following dictionaries:
d1 = "00f_5" :[1,2,3], "00f_6": [1,2,3]
d2 = ["marker":"00f_5",1: 'AAA',"marker":"00f_6", 1: 'CCC',"marker":"00f_5", 2:"AAC"]
I would like the following output:
d1 = "00f_5" :["AAA","AAC",3], "00f_6": ["CCC",2,3]
I have tried multiple attempts, but I couldn't get it. Any help would be much appreciated.
python python-3.x dictionary
python python-3.x dictionary
asked Mar 9 at 3:25
Mandie DriskillMandie Driskill
386
asked Mar 9 at 3:25
Mandie DriskillMandie Driskill
386
edited Mar 9 at 3:54
Mandie Driskill
asked Mar 9 at 3:25
Mandie DriskillMandie Driskill
386
asked Mar 9 at 3:25
Mandie DriskillMandie Driskill
386
asked Mar 9 at 3:25
Mandie DriskillMandie Driskill
386
386
1
Please post what you have attempted so far, and where did you encounter problems. [SO]: How to create a Minimal, Complete, and Verifiable example (mcve).
– CristiFati
Mar 9 at 3:29
add a comment |
1
Please post what you have attempted so far, and where did you encounter problems. [SO]: How to create a Minimal, Complete, and Verifiable example (mcve).
– CristiFati
Mar 9 at 3:29
1
1
Please post what you have attempted so far, and where did you encounter problems. [SO]: How to create a Minimal, Complete, and Verifiable example (mcve).
– CristiFati
Mar 9 at 3:29
Please post what you have attempted so far, and where did you encounter problems. [SO]: How to create a Minimal, Complete, and Verifiable example (mcve).
– CristiFati
Mar 9 at 3:29
add a comment |
3 Answers
3
active
oldest
votes
You can build a dict that maps the marker to a sub-dict that maps numeric keys to the 3-letter codes, so that given a marker and a numeric key, you can use dict.get method to get the mapped code if it exists:
d =
for s in d2:
d.setdefault(s['marker'], ).update(s)
d1 = k: [d[k].get(i, i) for i in l] for k, l in d1.items()
so that given:
d1 = "00f_5" :[1,2,3], "00f_6": [1,2,3]
d2 = ["marker":"00f_5",1: 'AAA',"marker":"00f_6", 1: 'CCC',"marker":"00f_5", 2:"AAC"]
d1 will become:
'00f_5': ['AAA', 'AAC', 3], '00f_6': ['CCC', 2, 3]
answered Mar 9 at 3:43
blhsingblhsing
42.6k41743
add a comment |
Although I will not give you the full program, you might try to understand and play with the following code:
d2 = ["marker":"00f_5",1: 'AAA',"marker":"00f_6", 1: 'CCC',"marker":"00f_5", 2:"AAC"]
for d in d2:
for key, value in d.items():
if key == "marker":
marker = value
else:
reference = value
content = key
print("Marker: , Reference: , Content: ".format(marker, reference, content))
answered Mar 9 at 3:37
mythenmetzmythenmetz
856
add a comment |
I would not suggest overwriting the first dict. Just make a new one (d3).
Given
d1 = "00f_5" :[1,2,3], "00f_6": [1,2,3]
lst = [
"marker":"00f_5",1: 'AAA',
"marker":"00f_6", 1: 'CCC',
"marker":"00f_5", 2:"AAC"
]
Code
d3 =
for d in lst:
key = d["marker"]
value = d1[key]
for k, v in d.items():
if not isinstance(k, int): # skip to numeric keys
continue
if key not in d3: # fill missing entries
d3[key] = value
idx = k - 1
d3[key][idx] = v # overwrite list item
d3
'00f_5': ['AAA', 'AAC', 3], '00f_6': ['CCC', 2, 3]
answered Mar 9 at 4:38
pylangpylang
14.4k24458
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can build a dict that maps the marker to a sub-dict that maps numeric keys to the 3-letter codes, so that given a marker and a numeric key, you can use dict.get method to get the mapped code if it exists:
d =
for s in d2:
d.setdefault(s['marker'], ).update(s)
d1 = k: [d[k].get(i, i) for i in l] for k, l in d1.items()
so that given:
d1 = "00f_5" :[1,2,3], "00f_6": [1,2,3]
d2 = ["marker":"00f_5",1: 'AAA',"marker":"00f_6", 1: 'CCC',"marker":"00f_5", 2:"AAC"]
d1 will become:
'00f_5': ['AAA', 'AAC', 3], '00f_6': ['CCC', 2, 3]
answered Mar 9 at 3:43
blhsingblhsing
42.6k41743
add a comment |
You can build a dict that maps the marker to a sub-dict that maps numeric keys to the 3-letter codes, so that given a marker and a numeric key, you can use dict.get method to get the mapped code if it exists:
d =
for s in d2:
d.setdefault(s['marker'], ).update(s)
d1 = k: [d[k].get(i, i) for i in l] for k, l in d1.items()
so that given:
d1 = "00f_5" :[1,2,3], "00f_6": [1,2,3]
d2 = ["marker":"00f_5",1: 'AAA',"marker":"00f_6", 1: 'CCC',"marker":"00f_5", 2:"AAC"]
d1 will become:
'00f_5': ['AAA', 'AAC', 3], '00f_6': ['CCC', 2, 3]
answered Mar 9 at 3:43
blhsingblhsing
42.6k41743
add a comment |
You can build a dict that maps the marker to a sub-dict that maps numeric keys to the 3-letter codes, so that given a marker and a numeric key, you can use dict.get method to get the mapped code if it exists:
d =
for s in d2:
d.setdefault(s['marker'], ).update(s)
d1 = k: [d[k].get(i, i) for i in l] for k, l in d1.items()
so that given:
d1 = "00f_5" :[1,2,3], "00f_6": [1,2,3]
d2 = ["marker":"00f_5",1: 'AAA',"marker":"00f_6", 1: 'CCC',"marker":"00f_5", 2:"AAC"]
d1 will become:
'00f_5': ['AAA', 'AAC', 3], '00f_6': ['CCC', 2, 3]
answered Mar 9 at 3:43
blhsingblhsing
42.6k41743
You can build a dict that maps the marker to a sub-dict that maps numeric keys to the 3-letter codes, so that given a marker and a numeric key, you can use dict.get method to get the mapped code if it exists:
d =
for s in d2:
d.setdefault(s['marker'], ).update(s)
d1 = k: [d[k].get(i, i) for i in l] for k, l in d1.items()
so that given:
d1 = "00f_5" :[1,2,3], "00f_6": [1,2,3]
d2 = ["marker":"00f_5",1: 'AAA',"marker":"00f_6", 1: 'CCC',"marker":"00f_5", 2:"AAC"]
d1 will become:
'00f_5': ['AAA', 'AAC', 3], '00f_6': ['CCC', 2, 3]
answered Mar 9 at 3:43
blhsingblhsing
42.6k41743
answered Mar 9 at 3:43
blhsingblhsing
42.6k41743
answered Mar 9 at 3:43
blhsingblhsing
42.6k41743
answered Mar 9 at 3:43
blhsingblhsing
42.6k41743
42.6k41743
add a comment |
add a comment |
Although I will not give you the full program, you might try to understand and play with the following code:
d2 = ["marker":"00f_5",1: 'AAA',"marker":"00f_6", 1: 'CCC',"marker":"00f_5", 2:"AAC"]
for d in d2:
for key, value in d.items():
if key == "marker":
marker = value
else:
reference = value
content = key
print("Marker: , Reference: , Content: ".format(marker, reference, content))
answered Mar 9 at 3:37
mythenmetzmythenmetz
856
add a comment |
Although I will not give you the full program, you might try to understand and play with the following code:
d2 = ["marker":"00f_5",1: 'AAA',"marker":"00f_6", 1: 'CCC',"marker":"00f_5", 2:"AAC"]
for d in d2:
for key, value in d.items():
if key == "marker":
marker = value
else:
reference = value
content = key
print("Marker: , Reference: , Content: ".format(marker, reference, content))
answered Mar 9 at 3:37
mythenmetzmythenmetz
856
add a comment |
Although I will not give you the full program, you might try to understand and play with the following code:
d2 = ["marker":"00f_5",1: 'AAA',"marker":"00f_6", 1: 'CCC',"marker":"00f_5", 2:"AAC"]
for d in d2:
for key, value in d.items():
if key == "marker":
marker = value
else:
reference = value
content = key
print("Marker: , Reference: , Content: ".format(marker, reference, content))
answered Mar 9 at 3:37
mythenmetzmythenmetz
856
Although I will not give you the full program, you might try to understand and play with the following code:
d2 = ["marker":"00f_5",1: 'AAA',"marker":"00f_6", 1: 'CCC',"marker":"00f_5", 2:"AAC"]
for d in d2:
for key, value in d.items():
if key == "marker":
marker = value
else:
reference = value
content = key
print("Marker: , Reference: , Content: ".format(marker, reference, content))
answered Mar 9 at 3:37
mythenmetzmythenmetz
856
answered Mar 9 at 3:37
mythenmetzmythenmetz
856
answered Mar 9 at 3:37
mythenmetzmythenmetz
856
answered Mar 9 at 3:37
mythenmetzmythenmetz
856
856
add a comment |
add a comment |
I would not suggest overwriting the first dict. Just make a new one (d3).
Given
d1 = "00f_5" :[1,2,3], "00f_6": [1,2,3]
lst = [
"marker":"00f_5",1: 'AAA',
"marker":"00f_6", 1: 'CCC',
"marker":"00f_5", 2:"AAC"
]
Code
d3 =
for d in lst:
key = d["marker"]
value = d1[key]
for k, v in d.items():
if not isinstance(k, int): # skip to numeric keys
continue
if key not in d3: # fill missing entries
d3[key] = value
idx = k - 1
d3[key][idx] = v # overwrite list item
d3
'00f_5': ['AAA', 'AAC', 3], '00f_6': ['CCC', 2, 3]
answered Mar 9 at 4:38
pylangpylang
14.4k24458
add a comment |
I would not suggest overwriting the first dict. Just make a new one (d3).
Given
d1 = "00f_5" :[1,2,3], "00f_6": [1,2,3]
lst = [
"marker":"00f_5",1: 'AAA',
"marker":"00f_6", 1: 'CCC',
"marker":"00f_5", 2:"AAC"
]
Code
d3 =
for d in lst:
key = d["marker"]
value = d1[key]
for k, v in d.items():
if not isinstance(k, int): # skip to numeric keys
continue
if key not in d3: # fill missing entries
d3[key] = value
idx = k - 1
d3[key][idx] = v # overwrite list item
d3
'00f_5': ['AAA', 'AAC', 3], '00f_6': ['CCC', 2, 3]
answered Mar 9 at 4:38
pylangpylang
14.4k24458
add a comment |
I would not suggest overwriting the first dict. Just make a new one (d3).
Given
d1 = "00f_5" :[1,2,3], "00f_6": [1,2,3]
lst = [
"marker":"00f_5",1: 'AAA',
"marker":"00f_6", 1: 'CCC',
"marker":"00f_5", 2:"AAC"
]
Code
d3 =
for d in lst:
key = d["marker"]
value = d1[key]
for k, v in d.items():
if not isinstance(k, int): # skip to numeric keys
continue
if key not in d3: # fill missing entries
d3[key] = value
idx = k - 1
d3[key][idx] = v # overwrite list item
d3
'00f_5': ['AAA', 'AAC', 3], '00f_6': ['CCC', 2, 3]
answered Mar 9 at 4:38
pylangpylang
14.4k24458
I would not suggest overwriting the first dict. Just make a new one (d3).
Given
d1 = "00f_5" :[1,2,3], "00f_6": [1,2,3]
lst = [
"marker":"00f_5",1: 'AAA',
"marker":"00f_6", 1: 'CCC',
"marker":"00f_5", 2:"AAC"
]
Code
d3 =
for d in lst:
key = d["marker"]
value = d1[key]
for k, v in d.items():
if not isinstance(k, int): # skip to numeric keys
continue
if key not in d3: # fill missing entries
d3[key] = value
idx = k - 1
d3[key][idx] = v # overwrite list item
d3
'00f_5': ['AAA', 'AAC', 3], '00f_6': ['CCC', 2, 3]
answered Mar 9 at 4:38
pylangpylang
14.4k24458
answered Mar 9 at 4:38
pylangpylang
14.4k24458
answered Mar 9 at 4:38
pylangpylang
14.4k24458
answered Mar 9 at 4:38
pylangpylang
14.4k24458
14.4k24458
add a comment |
add a comment |
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1
Please post what you have attempted so far, and where did you encounter problems. [SO]: How to create a Minimal, Complete, and Verifiable example (mcve).
– CristiFati
Mar 9 at 3:29