I have written a program which gives me the following outputs for five nodes which is the shortest path from each node to different nodes :
G1= {'D': 3.0, 'E': 4.0, 'B': 1.0, 'C': 5.0, 'A': 0}
G1={'D': 2.0, 'E': 3.0, 'B': 0, 'C': 4.0, 'A': 1.0}
G1={'D': 2.0, 'E': 3.0, 'B': 4.0, 'C': 0, 'A': 5.0}
G1={'D': 0, 'E': 1.0, 'B': 2.0, 'C': 2.0, 'A': 3.0}
G1={'D': 1.0, 'E': 0, 'B': 3.0, 'C': 3.0, 'A': 4.0}
I am trying to find the mean of all of the nodes from the above output. I tried the following code :
for s in G:
G1=ShortestPaths(G,s)#this gives the output i mentioned above
mean= sum([G1[s] for s in G1])/(len(G1)-1)# this is where i am not getting result
return float(mean)
But it is giving mean of only the last line.I need sum of all the values in the dictionary(sum of 25 values) and divide by 20(since there is a zero in every line of my output.I should not consider that). Can anyone help me with this with a simple code?? I am not suppose to .items and other built-in functions.
Calculate the mean at the end, after the loop:
total = 0.0
count = 0.0
for s in G:
G1=ShortestPaths(G,s)
total += sum([G1[s] for s in G1])
count += (len(G1)-1)
return float(total / count) if count else None
Related
I have checked the advicse here: Nested dictionary to multiindex dataframe where dictionary keys are column labels
However, I couldn't get it to work in my problem.
I would like to change a dictionary into multiindexed dataframe, where 'a','b','c' are names of multiindexes, their values 12,0.8,1.8,bla1,bla2,bla3,bla4 are multiindexes and values from lists are assign to the multiindexes as in the picture of table below.
My dictionary:
dictionary ={
"{'a': 12.0, 'b': 0.8, 'c': ' bla1'}": [200, 0.0, '0.0'],
"{'a': 12.0, 'b': 0.8, 'c': ' bla2'}": [37, 44, '0.6'],
"{'a': 12.0, 'b': 1.8, 'c': ' bla3'}": [100, 2.0, '1.0'],
"{'a': 12.0, 'b': 1.8, 'c': ' bla4'}": [400, 3.0, '1.0']
}
The result DataFrame I would like to get:
The code which don't make multiindexes and set every values under each other in next row:
df_a = pd.DataFrame.from_dict(dictionary, orient="index").stack().to_frame()
df_b = pd.DataFrame(df_a[0].values.tolist(), index=df_a.index)
Use ast.literal_eval to convert each string into a dictionary and build the index from there:
import pandas as pd
from ast import literal_eval
dictionary ={
"{'a': 12.0, 'b': 0.8, 'c': ' bla1'}": [200, 0.0, '0.0'],
"{'a': 12.0, 'b': 0.8, 'c': ' bla2'}": [37, 44, '0.6'],
"{'a': 12.0, 'b': 1.8, 'c': ' bla3'}": [100, 2.0, '1.0'],
"{'a': 12.0, 'b': 1.8, 'c': ' bla4'}": [400, 3.0, '1.0']
}
keys, data = zip(*dictionary.items())
index = pd.MultiIndex.from_frame(pd.DataFrame([literal_eval(i) for i in keys]))
res = pd.DataFrame(data=list(data), index=index)
print(res)
Output
0 1 2
a b c
12.0 0.8 bla1 200 0.0 0.0
bla2 37 44.0 0.6
1.8 bla3 100 2.0 1.0
bla4 400 3.0 1.0
I have a nested dictionary, such as:
{'A1': {'T1': [1, 3.0, 3, 4.0], 'T2': [2, 2.0]}, 'A2': {'T1': [1, 0.0, 3, 5.0], 'T2': [2, 3.0]}}
What I want to do is sum each sub dictionary, to obtain this:
A1 A2 A1 A2
T1+T1 T2+T2 (ignore the first entry of the list)
[3.0, 5.0, 9.0] <<<< output
1 2 3
res 3.0 + 0.0 = 3.0 and 2.0 + 3.0 = 5.0 and 5.0 + 4.0 = 9.0
How can I do this? I've tried a for, but I've created a big mess
One way is to use collections.Counter in a list comprehension, and sum the resulting Counter objects:
from collections import Counter
d = {'A1': {'T1': 3.0, 'T2': 2.0}, 'A2': {'T1': 0.0, 'T2': 3.0}}
l = (Counter(i) for i in d.values())
sum(l, Counter())
# Counter({'T1': 3.0, 'T2': 5.0})
For sum to work here, I've defined an empty Counter() as the start argument, so sum expects other Counter objects.
To get only the values, you can do:
sum(l, Counter()).values()
# dict_values([3.0, 5.0])
you could use a list comprehension with zip:
d = {'A1': {'T1': 3.0, 'T2': 2.0}, 'A2': {'T1': 0.0, 'T2': 3.0}}
[sum(e) for e in zip(*(e.values() for e in d.values()))]
output:
[3.0, 5.0]
this will work if your python version is >= 3.6
also, you can use 2 for loops:
r = {}
for dv in d.values():
for k, v in dv.items():
r.setdefault(k, []).append(v)
result = [sum(v) for v in r.values()]
print(result)
output:
[3.0, 5.0]
after your edit
you could use:
from itertools import zip_longest
sum_t1, sum_t2 = list(list(map(sum, zip(*t))) for t in zip(*[e.values() for e in d.values()]))
[i for t in zip_longest(sum_t1[1:], sum_t2[1:]) for i in t if i is not None]
output:
[3.0, 5.0, 6, 9.0]
I have a data frame with three columns, I would like to create a dictionary after applying groupby function on first and second column.I can do this by for loops, but is there any pandas way of doing it?
DataFrame:
Col X Col Y Sum
A a 3
A b 2
A c 1
B p 5
B q 6
B r 7
After grouping by on Col X and Col Y : df.groupby(['Col X','Col Y']).sum()
Sum
Col X Col Y
A a 3
b 2
c 1
B p 5
q 6
r 7
Dictionary I want to create
{A:{'a':3,'b':2,'c':1}, B:{'p':5,'q':6,'r':7}}
Use a dictionary comprehension while iterating via a groupby object
{name: dict(zip(g['Col Y'], g['Sum'])) for name, g in df.groupby('Col X')}
{'A': {'a': 3, 'b': 2, 'c': 1}, 'B': {'p': 5, 'q': 6, 'r': 7}}
If you insisted on using to_dict somewhere, you could do something like this:
s = df.set_index(['Col X', 'Col Y']).Sum
{k: s.xs(k).to_dict() for k in s.index.levels[0]}
{'A': {'a': 3, 'b': 2, 'c': 1}, 'B': {'p': 5, 'q': 6, 'r': 7}}
Keep in mind, that the to_dict method is just using some comprehension under the hood. If you have a special use case that requires something more than what the orient options provide for... there is no shame in constructing your own comprehension.
You can iterate over the MultiIndex series:
>>> s = df.set_index(['ColX', 'ColY'])['Sum']
>>> {k: v.reset_index(level=0, drop=True).to_dict() for k, v in s.groupby(level=0)}
{'A': {'a': 3, 'b': 2, 'c': 1}, 'B': {'p': 5, 'q': 6, 'r': 7}}
#A to_dict() solution
d = df.groupby(['Col X','Col Y']).sum().reset_index().pivot(columns='Col X',values='Sum').to_dict()
Out[70]:
{'A': {0: 3.0, 1: 2.0, 2: 1.0, 3: nan, 4: nan, 5: nan},
'B': {0: nan, 1: nan, 2: nan, 3: 5.0, 4: 6.0, 5: 7.0}}
#if you need to get rid of the nans:
{k1:{k2:v2 for k2,v2 in v1.items() if pd.notnull(v2)} for k1,v1 in d.items()}
Out[73]: {'A': {0: 3.0, 1: 2.0, 2: 1.0}, 'B': {3: 5.0, 4: 6.0, 5: 7.0}}
I have a following dictionary:
centroid = {'A': [1.0, 1.0], 'B': [2.0, 1.0]}
Using the above dictionary I am creating two different dictionaries and appending them to a list:
for key in centroids:
clusters_list.append(dict(zip(key, centroids.get(key))))
However when I check my cluster_list I get the following data:
[{'A': 1.0}, {'B': 2.0}]
instead of
[{'A': [1.0, 1.0]}, {'B': [2.0, 1.0]}].
How can i fix this?
You can use a list comprehension:
For Python 2:
cluster_list = [{k: v} for k, v in centroid.iteritems()]
# [{'A': [1.0, 1.0]}, {'B': [2.0, 1.0]}]
For Python 3:
cluster_list = [{k: v} for k, v in centroid.items()]
You can also use starmap from itertools module.
In [1]: from itertools import starmap
In [2]: list(starmap(lambda k,v: {k:v}, centroid.items()))
Out[2]: [{'B': [2.0, 1.0]}, {'A': [1.0, 1.0]}]
And of course, it doesn't guarantee the order in the resulting list.
This is a follow-up question to this and I looked at this one as well. My Python script collects data during every run and sends data to a URL. The URL should display data/values up to 2 decimal places. I modified my script to round off the values to 2 decimal places as shown below. After doing so, any and all print statements show that my script is rounding off to 2 decimal places. Even after urlencode, when I decode the params, it shows 2 decimal places. But at the URL, the values show up with all the decimal places - instead of 1.0, say, it shows 1.000342760436734 whereas the values should show as 1.00 and not more.
Here's the code in my script with the print outputs:
//The data that is collected by the script - a list of dicts.
y = [{'a': 80.0, 'b': 0.0786235, 'c': 10.0, 'd': 10.6742903}, {'a': 80.73246, 'b': 0.0, 'c':
10.780323, 'd': 10.0}, {'a': 80.7239, 'b': 0.7823640, 'c': 10.0, 'd': 10.0}, {'a':
80.7802313217234, 'b': 0.0, 'c': 10.0, 'd': 10.9762304}]
//The code that rounds off the decimal places to 2
class LessPrecise(float):
def __repr__(self):
return str(self)
def roundingVals_toTwoDeci(y):
for d in y:
for k, v in d.iteritems():
v = LessPrecise(round(v, 2))
print v
d[k] = v
//Json.dumps
roundingVals_toTwoDeci(y)
j = json.dumps(y)
print j
//print j gives
[{"a": 80.0, "b": 0.0, "c": 10.0, "d": 10.0}, {"a": 100.0, "b": 0.0, "c": 0.0, "d": 0.0}, {"a":
80.0, "b": 0.0, "c": 10.0, "d": 10.0}, {"a": 90.0, "b": 0.0, "c": 0.0, "d": 10.0}]
//urlencoding it
params = urllib.urlencode({'thekey': j})
//after decoding params, I get
thekey=[{"a": 80.0, "b": 0.0, "c": 10.0, "d": 10.0}, {"a": 100.0, "b": 0.0, "c": 0.0, "d":
0.0}, {"a": 80.0, "b": 0.0, "c": 10.0, "d": 10.0}, {"a": 90.0, "b": 0.0, "c": 0.0, "d": 10.0}]
//So far, so good. At the URL, however, instead of 10.0, it shows 10.000436783313897. I don't what's going on or how to fix it.
I should mention that I get the same float values at the URL even after I convert values into strings directly as in:
def roundingVals_toTwoDeci(y):
for d in y:
for k, v in d.iteritems():
d[k] = str(round(v, 2))
return
With this, print s=json.dumps() gives values like {"a": "10.0", "b": "3.1", etc.} but at the URL, values are 10.856473985798743.