How to add multiple values to a single key
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In my code I have generated two lists, t_fail and p_fail. For these two lists the index number relates both p_fail and t_fail. So for the 100 in t_fail, the 100 in p_fail corresponds to it.
t_fail = [100,100,200,300]
p_fail = [100,105,105,110]
I want to make a dictionary that has the t_fail values as keys and the p_fail values as the values of the dictionary. But if the same t_fail value repeats itself, I want to be able to append to the corresponding p_fail value to the dictionary key.
for example:
dict = {
100: [100,105]
200: [105]
300: [110]
}
I was thinking about using some sort of for loop but other than that, I'm a little lost.
Thank you
python-3.x
asked Jan 4 at 16:04
bbalzanobbalzano
185
add a comment |
In my code I have generated two lists, t_fail and p_fail. For these two lists the index number relates both p_fail and t_fail. So for the 100 in t_fail, the 100 in p_fail corresponds to it.
t_fail = [100,100,200,300]
p_fail = [100,105,105,110]
I want to make a dictionary that has the t_fail values as keys and the p_fail values as the values of the dictionary. But if the same t_fail value repeats itself, I want to be able to append to the corresponding p_fail value to the dictionary key.
for example:
dict = {
100: [100,105]
200: [105]
300: [110]
}
I was thinking about using some sort of for loop but other than that, I'm a little lost.
Thank you
python-3.x
asked Jan 4 at 16:04
bbalzanobbalzano
185
add a comment |
In my code I have generated two lists, t_fail and p_fail. For these two lists the index number relates both p_fail and t_fail. So for the 100 in t_fail, the 100 in p_fail corresponds to it.
t_fail = [100,100,200,300]
p_fail = [100,105,105,110]
I want to make a dictionary that has the t_fail values as keys and the p_fail values as the values of the dictionary. But if the same t_fail value repeats itself, I want to be able to append to the corresponding p_fail value to the dictionary key.
for example:
dict = {
100: [100,105]
200: [105]
300: [110]
}
I was thinking about using some sort of for loop but other than that, I'm a little lost.
Thank you
python-3.x
asked Jan 4 at 16:04
bbalzanobbalzano
185
In my code I have generated two lists, t_fail and p_fail. For these two lists the index number relates both p_fail and t_fail. So for the 100 in t_fail, the 100 in p_fail corresponds to it.
t_fail = [100,100,200,300]
p_fail = [100,105,105,110]
I want to make a dictionary that has the t_fail values as keys and the p_fail values as the values of the dictionary. But if the same t_fail value repeats itself, I want to be able to append to the corresponding p_fail value to the dictionary key.
for example:
dict = {
100: [100,105]
200: [105]
300: [110]
}
I was thinking about using some sort of for loop but other than that, I'm a little lost.
Thank you
python-3.x
python-3.x
asked Jan 4 at 16:04
bbalzanobbalzano
185
asked Jan 4 at 16:04
bbalzanobbalzano
185
asked Jan 4 at 16:04
bbalzanobbalzano
185
asked Jan 4 at 16:04
bbalzanobbalzano
185
asked Jan 4 at 16:04
bbalzanobbalzano
185
185
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
As mentionned @Marcus, this is a good use case for defaultdict
from collections import defaultdict
t_fail = [100,100,200,300]
p_fail = [100,105,105,110]
d = defaultdict(list)
for t, p in zip(t_fail, p_fail):
d[t].append(p)
print(d)
# defaultdict(<class 'list'>, {100: [100, 105], 200: [105], 300: [110]})
Two ways to pretty print your dict
First by looping over the items
for k, v in sorted(d.items()):
print(k,v)
The second, by unpacking the value and specifying the separator
print(*sorted(d.items()), sep='n')
answered Jan 4 at 16:29
BlueSheepTokenBlueSheepToken
2,1691617
Is it possible for it to print without the "defaultdict(<class 'list'>," section and have each key value pair on it's own line?
– bbalzano
Jan 4 at 17:11
@bbalzano I edited my answer, to fillful your needs
– BlueSheepToken
Jan 4 at 17:17
Is there any way to save the sorted version of the dictionary to a text file? The only way I have been able to save it to a text file is to save the original version
– bbalzano
Jan 4 at 19:57
You can dump it into a json file
– BlueSheepToken
Jan 4 at 19:58
You might want to open a new question for thius :)
– BlueSheepToken
Jan 4 at 19:59
add a comment |
You can try like this.
As dictionary is an unordered data type so if you are looking for ordered keys, you can use OrderedDict from collections module.
d = {} # d = dict()
for key, value in zip(t_fail, p_fail):
if key in d:
d[key].append(value)
else:
d[key] = [value]
print(d) # {200: [105], 300: [110], 100: [100, 105]}
Pretty printing the dictionary.
import json
s = json.dumps(d, indent=4)
print(s)
# {
# "200": [
# 105
# ],
# "300": [
# 110
# ],
# "100": [
# 100,
# 105
# ]
# }
answered Jan 4 at 16:13
hygullhygull
4,05921632
This is a good use case for a defaultdict. The example in the documentation shows how it is used with lists: docs.python.org/3.7/library/…
– Marcus
Jan 4 at 16:16
Yeah, that is excellent. Thank you. I know it, it is just to use the concept of dictionary instead of using the already implemented feature.
– hygull
Jan 4 at 16:18
What's the relevance of OrderedDict here?
– glibdud
Jan 4 at 16:20
I do not see any use but the problem has o/p with keys in sequence as they appear in the list. So, I thought, user will be looking for keys in order also and that forced me to update.
– hygull
Jan 4 at 16:25
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
As mentionned @Marcus, this is a good use case for defaultdict
from collections import defaultdict
t_fail = [100,100,200,300]
p_fail = [100,105,105,110]
d = defaultdict(list)
for t, p in zip(t_fail, p_fail):
d[t].append(p)
print(d)
# defaultdict(<class 'list'>, {100: [100, 105], 200: [105], 300: [110]})
Two ways to pretty print your dict
First by looping over the items
for k, v in sorted(d.items()):
print(k,v)
The second, by unpacking the value and specifying the separator
print(*sorted(d.items()), sep='n')
answered Jan 4 at 16:29
BlueSheepTokenBlueSheepToken
2,1691617
Is it possible for it to print without the "defaultdict(<class 'list'>," section and have each key value pair on it's own line?
– bbalzano
Jan 4 at 17:11
@bbalzano I edited my answer, to fillful your needs
– BlueSheepToken
Jan 4 at 17:17
Is there any way to save the sorted version of the dictionary to a text file? The only way I have been able to save it to a text file is to save the original version
– bbalzano
Jan 4 at 19:57
You can dump it into a json file
– BlueSheepToken
Jan 4 at 19:58
You might want to open a new question for thius :)
– BlueSheepToken
Jan 4 at 19:59
add a comment |
As mentionned @Marcus, this is a good use case for defaultdict
from collections import defaultdict
t_fail = [100,100,200,300]
p_fail = [100,105,105,110]
d = defaultdict(list)
for t, p in zip(t_fail, p_fail):
d[t].append(p)
print(d)
# defaultdict(<class 'list'>, {100: [100, 105], 200: [105], 300: [110]})
Two ways to pretty print your dict
First by looping over the items
for k, v in sorted(d.items()):
print(k,v)
The second, by unpacking the value and specifying the separator
print(*sorted(d.items()), sep='n')
answered Jan 4 at 16:29
BlueSheepTokenBlueSheepToken
2,1691617
Is it possible for it to print without the "defaultdict(<class 'list'>," section and have each key value pair on it's own line?
– bbalzano
Jan 4 at 17:11
@bbalzano I edited my answer, to fillful your needs
– BlueSheepToken
Jan 4 at 17:17
Is there any way to save the sorted version of the dictionary to a text file? The only way I have been able to save it to a text file is to save the original version
– bbalzano
Jan 4 at 19:57
You can dump it into a json file
– BlueSheepToken
Jan 4 at 19:58
You might want to open a new question for thius :)
– BlueSheepToken
Jan 4 at 19:59
add a comment |
As mentionned @Marcus, this is a good use case for defaultdict
from collections import defaultdict
t_fail = [100,100,200,300]
p_fail = [100,105,105,110]
d = defaultdict(list)
for t, p in zip(t_fail, p_fail):
d[t].append(p)
print(d)
# defaultdict(<class 'list'>, {100: [100, 105], 200: [105], 300: [110]})
Two ways to pretty print your dict
First by looping over the items
for k, v in sorted(d.items()):
print(k,v)
The second, by unpacking the value and specifying the separator
print(*sorted(d.items()), sep='n')
answered Jan 4 at 16:29
BlueSheepTokenBlueSheepToken
2,1691617
As mentionned @Marcus, this is a good use case for defaultdict
from collections import defaultdict
t_fail = [100,100,200,300]
p_fail = [100,105,105,110]
d = defaultdict(list)
for t, p in zip(t_fail, p_fail):
d[t].append(p)
print(d)
# defaultdict(<class 'list'>, {100: [100, 105], 200: [105], 300: [110]})
Two ways to pretty print your dict
First by looping over the items
for k, v in sorted(d.items()):
print(k,v)
The second, by unpacking the value and specifying the separator
print(*sorted(d.items()), sep='n')
answered Jan 4 at 16:29
BlueSheepTokenBlueSheepToken
2,1691617
edited Jan 4 at 17:17
answered Jan 4 at 16:29
BlueSheepTokenBlueSheepToken
2,1691617
answered Jan 4 at 16:29
BlueSheepTokenBlueSheepToken
2,1691617
answered Jan 4 at 16:29
BlueSheepTokenBlueSheepToken
2,1691617
2,1691617
Is it possible for it to print without the "defaultdict(<class 'list'>," section and have each key value pair on it's own line?
– bbalzano
Jan 4 at 17:11
@bbalzano I edited my answer, to fillful your needs
– BlueSheepToken
Jan 4 at 17:17
Is there any way to save the sorted version of the dictionary to a text file? The only way I have been able to save it to a text file is to save the original version
– bbalzano
Jan 4 at 19:57
You can dump it into a json file
– BlueSheepToken
Jan 4 at 19:58
You might want to open a new question for thius :)
– BlueSheepToken
Jan 4 at 19:59
add a comment |
Is it possible for it to print without the "defaultdict(<class 'list'>," section and have each key value pair on it's own line?
– bbalzano
Jan 4 at 17:11
@bbalzano I edited my answer, to fillful your needs
– BlueSheepToken
Jan 4 at 17:17
Is there any way to save the sorted version of the dictionary to a text file? The only way I have been able to save it to a text file is to save the original version
– bbalzano
Jan 4 at 19:57
You can dump it into a json file
– BlueSheepToken
Jan 4 at 19:58
You might want to open a new question for thius :)
– BlueSheepToken
Jan 4 at 19:59
Is it possible for it to print without the "defaultdict(<class 'list'>," section and have each key value pair on it's own line?
– bbalzano
Jan 4 at 17:11
Is it possible for it to print without the "defaultdict(<class 'list'>," section and have each key value pair on it's own line?
– bbalzano
Jan 4 at 17:11
@bbalzano I edited my answer, to fillful your needs
– BlueSheepToken
Jan 4 at 17:17
@bbalzano I edited my answer, to fillful your needs
– BlueSheepToken
Jan 4 at 17:17
Is there any way to save the sorted version of the dictionary to a text file? The only way I have been able to save it to a text file is to save the original version
– bbalzano
Jan 4 at 19:57
Is there any way to save the sorted version of the dictionary to a text file? The only way I have been able to save it to a text file is to save the original version
– bbalzano
Jan 4 at 19:57
You can dump it into a json file
– BlueSheepToken
Jan 4 at 19:58
You can dump it into a json file
– BlueSheepToken
Jan 4 at 19:58
You might want to open a new question for thius :)
– BlueSheepToken
Jan 4 at 19:59
You might want to open a new question for thius :)
– BlueSheepToken
Jan 4 at 19:59
add a comment |
You can try like this.
As dictionary is an unordered data type so if you are looking for ordered keys, you can use OrderedDict from collections module.
d = {} # d = dict()
for key, value in zip(t_fail, p_fail):
if key in d:
d[key].append(value)
else:
d[key] = [value]
print(d) # {200: [105], 300: [110], 100: [100, 105]}
Pretty printing the dictionary.
import json
s = json.dumps(d, indent=4)
print(s)
# {
# "200": [
# 105
# ],
# "300": [
# 110
# ],
# "100": [
# 100,
# 105
# ]
# }
answered Jan 4 at 16:13
hygullhygull
4,05921632
This is a good use case for a defaultdict. The example in the documentation shows how it is used with lists: docs.python.org/3.7/library/…
– Marcus
Jan 4 at 16:16
Yeah, that is excellent. Thank you. I know it, it is just to use the concept of dictionary instead of using the already implemented feature.
– hygull
Jan 4 at 16:18
What's the relevance of OrderedDict here?
– glibdud
Jan 4 at 16:20
I do not see any use but the problem has o/p with keys in sequence as they appear in the list. So, I thought, user will be looking for keys in order also and that forced me to update.
– hygull
Jan 4 at 16:25
add a comment |
You can try like this.
As dictionary is an unordered data type so if you are looking for ordered keys, you can use OrderedDict from collections module.
d = {} # d = dict()
for key, value in zip(t_fail, p_fail):
if key in d:
d[key].append(value)
else:
d[key] = [value]
print(d) # {200: [105], 300: [110], 100: [100, 105]}
Pretty printing the dictionary.
import json
s = json.dumps(d, indent=4)
print(s)
# {
# "200": [
# 105
# ],
# "300": [
# 110
# ],
# "100": [
# 100,
# 105
# ]
# }
answered Jan 4 at 16:13
hygullhygull
4,05921632
This is a good use case for a defaultdict. The example in the documentation shows how it is used with lists: docs.python.org/3.7/library/…
– Marcus
Jan 4 at 16:16
Yeah, that is excellent. Thank you. I know it, it is just to use the concept of dictionary instead of using the already implemented feature.
– hygull
Jan 4 at 16:18
What's the relevance of OrderedDict here?
– glibdud
Jan 4 at 16:20
I do not see any use but the problem has o/p with keys in sequence as they appear in the list. So, I thought, user will be looking for keys in order also and that forced me to update.
– hygull
Jan 4 at 16:25
add a comment |
You can try like this.
As dictionary is an unordered data type so if you are looking for ordered keys, you can use OrderedDict from collections module.
d = {} # d = dict()
for key, value in zip(t_fail, p_fail):
if key in d:
d[key].append(value)
else:
d[key] = [value]
print(d) # {200: [105], 300: [110], 100: [100, 105]}
Pretty printing the dictionary.
import json
s = json.dumps(d, indent=4)
print(s)
# {
# "200": [
# 105
# ],
# "300": [
# 110
# ],
# "100": [
# 100,
# 105
# ]
# }
answered Jan 4 at 16:13
hygullhygull
4,05921632
You can try like this.
As dictionary is an unordered data type so if you are looking for ordered keys, you can use OrderedDict from collections module.
d = {} # d = dict()
for key, value in zip(t_fail, p_fail):
if key in d:
d[key].append(value)
else:
d[key] = [value]
print(d) # {200: [105], 300: [110], 100: [100, 105]}
Pretty printing the dictionary.
import json
s = json.dumps(d, indent=4)
print(s)
# {
# "200": [
# 105
# ],
# "300": [
# 110
# ],
# "100": [
# 100,
# 105
# ]
# }
answered Jan 4 at 16:13
hygullhygull
4,05921632
edited Jan 4 at 16:23
answered Jan 4 at 16:13
hygullhygull
4,05921632
answered Jan 4 at 16:13
hygullhygull
4,05921632
answered Jan 4 at 16:13
hygullhygull
4,05921632
4,05921632
This is a good use case for a defaultdict. The example in the documentation shows how it is used with lists: docs.python.org/3.7/library/…
– Marcus
Jan 4 at 16:16
Yeah, that is excellent. Thank you. I know it, it is just to use the concept of dictionary instead of using the already implemented feature.
– hygull
Jan 4 at 16:18
What's the relevance of OrderedDict here?
– glibdud
Jan 4 at 16:20
I do not see any use but the problem has o/p with keys in sequence as they appear in the list. So, I thought, user will be looking for keys in order also and that forced me to update.
– hygull
Jan 4 at 16:25
add a comment |
This is a good use case for a defaultdict. The example in the documentation shows how it is used with lists: docs.python.org/3.7/library/…
– Marcus
Jan 4 at 16:16
Yeah, that is excellent. Thank you. I know it, it is just to use the concept of dictionary instead of using the already implemented feature.
– hygull
Jan 4 at 16:18
What's the relevance of OrderedDict here?
– glibdud
Jan 4 at 16:20
I do not see any use but the problem has o/p with keys in sequence as they appear in the list. So, I thought, user will be looking for keys in order also and that forced me to update.
– hygull
Jan 4 at 16:25
This is a good use case for a defaultdict. The example in the documentation shows how it is used with lists: docs.python.org/3.7/library/…
– Marcus
Jan 4 at 16:16
This is a good use case for a defaultdict. The example in the documentation shows how it is used with lists: docs.python.org/3.7/library/…
– Marcus
Jan 4 at 16:16
Yeah, that is excellent. Thank you. I know it, it is just to use the concept of dictionary instead of using the already implemented feature.
– hygull
Jan 4 at 16:18
Yeah, that is excellent. Thank you. I know it, it is just to use the concept of dictionary instead of using the already implemented feature.
– hygull
Jan 4 at 16:18
What's the relevance of OrderedDict here?
– glibdud
Jan 4 at 16:20
What's the relevance of OrderedDict here?
– glibdud
Jan 4 at 16:20
I do not see any use but the problem has o/p with keys in sequence as they appear in the list. So, I thought, user will be looking for keys in order also and that forced me to update.
– hygull
Jan 4 at 16:25
I do not see any use but the problem has o/p with keys in sequence as they appear in the list. So, I thought, user will be looking for keys in order also and that forced me to update.
– hygull
Jan 4 at 16:25
add a comment |
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