If I have a dictionary like this, filled with similar lists, how can I apply a while loo tp extract a list that prints that second element:
racoona_valence={}
racoona_valence={"rs13283416": ["7:87345874365-839479328749+","BOBB7"],\}
I need to print the part that says "BOBB7" for 2nd element of the lists in a larger dictionary. There are ten key-value pairs in it, so I am starting it like so, but unsure what to do because all the examples I can find don't relate to my problem:
n=10
gene_list = []
while n>0:
Any help greatly appreciated.
Well, there's a bunch of ways to do it depending on how well-structured your data is.
racoona_valence={"rs13283416": ["7:87345874365-839479328749+","BOBB7"], "rs13283414": ["7:87345874365-839479328749+","BOBB4"]}
output = []
for key in racoona_valence.keys():
output.append(racoona_valence[key][1])
print(output)
other_output = []
for key, value in racoona_valence.items():
other_output.append(value[1])
print(other_output)
list_comprehension = [value[1] for value in racoona_valence.values()]
print(list_comprehension)
n = len(racoona_valence.values())-1
counter = 0
gene_list = []
while counter<=n:
gene_list.append(list(racoona_valence.values())[n][1])
counter += 1
print(gene_list)
Here is a list comprehension that does what you want:
second_element = [x[1] for x in racoona_valence.values()]
Here is a for loop that does what you want:
second_element = []
for value in racoona_valence.values():
second_element.append(value[1])
Here is a while loop that does what you want:
# don't use a while loop to loop over iterables, it's a bad idea
i = 0
second_element = []
dict_values = list(racoona_valence.values())
while i < len(dict_values):
second_element.append(dict_values[i][1])
i += 1
Regardless of which approach you use, you can see the results by doing the following:
for item in second_element:
print(item)
For the example that you gave, this is the output:
BOBB7
How do I code a function in python which can:
iterate through a list of word strings which may contain duplicate words and referencing to a dictionary,
find the word with the highest absolute sum, and
output it along with the corresponding absolute value.
The function also has to ignore words which are not in the dictionary.
For example,
Assume the function is called H_abs_W().
Given the following list and dict:
list_1 = ['apples','oranges','pears','apples']
Dict_1 = {'apples':5.23,'pears':-7.62}
Then calling the function as:
H_abs_W(list_1,Dict_1)
Should give the output:
'apples',10.46
EDIT:
I managed to do it in the end with the code below. Looking over the answers, turns out I could have done it in a shorter fashion, lol.
def H_abs_W(list_1,Dict_1):
freqW = {}
for char in list_1:
if char in freqW:
freqW[char] += 1
else:
freqW[char] = 1
ASum_W = 0
i_word = ''
for a,b in freqW.items():
x = 0
d = Dict_1.get(a,0)
x = abs(float(b)*float(d))
if x > ASum_W:
ASum_W = x
i_word = a
return(i_word,ASum_W)
list_1 = ['apples','oranges','pears','apples']
Dict_1 = {'apples':5.23,'pears':-7.62}
d = {k:0 for k in list_1}
for x in list_1:
if x in Dict_1.keys():
d[x]+=Dict_1[x]
m = max(Dict_1, key=Dict_1.get)
print(m,Dict_1[m])
try this,
key, value = sorted(Dict_1.items(), key = lambda x : x[1], reverse=True)[0]
print(f"{key}, {list_1.count(key) * value}")
# apples, 10.46
you can use Counter to calculate the frequency(number of occurrences) of each item in the list.
max(counter.values()) will give us the count of maximum occurring element
max(counter, key=counter.get) will give the which item in the list is
associated with that highest count.
========================================================================
from collections import Counter
def H_abs_W(list_1, Dict_1):
counter = Counter(list_1)
count = max(counter.values())
item = max(counter, key=counter.get)
return item, abs(count * Dict_1.get(item))
I am quite new to python so still getting to grips with the language.
I have the following function which takes a string and apply it to an algorithm which tells us if it aligns to models 1, 2, 3, 4, or 5.
Currently this piece of code:
def apply_text(text):
test_str = [text]
test_new = tfidf_m.transform(test_str)
prediction = 0
for m in range(0,5):
percentage = '{P:.1%}'.format(M=cat[m], P=lr_m[m].predict_proba(test_new)[0][1])
print(percentage)
And running the following function: apply_text('Terrible idea.')
Gives the following output:
71.4%
33.1%
2.9%
1.6%
4.9%
With Model 1 = 71.4%, Model 2 = 33.1%, ... Model 5 = 4.9%.
I want to only output the Model number where there is the highest percentage. So in the above example, the answer would be 1 as this has 71.4%.
As the output is a string type I am finding it difficult to find ways of converting this to an int and then comparing each value (probably in a loop of some sort) to obtain the maximum value
I think you want to save the percentages along with the model number, sort it and then return the highest.
This can be done by something like this:
def apply_text(text):
test_str = [text]
test_new = tfidf_m.transform(test_str)
prediction = 0
percentage_list = []
for m in range(0,5):
percentage = '{P:.1}'.format(M=cat[m], P=lr_m[m].predict_proba(test_new)[0][1])
percentage_list.append([m+1, float(percentage)])
percentage_list.sort(reverse=True, key=lambda a: a[1])
return percentage_list[0][0]
Things to note:
Sorting in reverse order as default is ascending. You could skip reversing and access the last element of precentage_list by accessing -1 element
The key function is used as we need to sort using the percentage
Try putting values in a list then you can utilize list methods:
percentage = []
for m in range(0, 5):
percentage.append('{P:.1%}'.format(M=cat[m], P=lr_m[m].predict_proba(test_new)[0][1]))
print(*percentage, sep='\n')
print('Max on model', percentage.index(max(percentage)))
Or using a dictionary:
percentage = {}
for m in range(0, 5):
percentage['Model ' + str(m)] = '{P:.1%}'.format(M=cat[m], P=lr_m[m].predict_proba(test_new)[0][1])
print(*percentage, sep='\n')
print('Max on', max(percentage.keys(), key=(lambda key: percentage[key])))
I have a dictionary with each key containing multiple values (list):
one = [1,2,3]
two = [1,2,3]
three= [1,2,3]
It was obtained with the following line of code:
output_file.write('{0}\t{1}\n'.format(key,"\t".join(value)))
So my final printed output looks like this:
one 1 2 3
two 1 2 3
three 1 2 3
My goal now is to have the output looking like this instead:
one 1
one 2
one 3
two 1
two 2
…
Any tips?
You can add the key itself as delimiter
#key = "one"
#value = ['1','2','3']
print(key+'\t'+'\n{0}\t'.format(key).join(value))
output
one 1
one 2
one 3
You could do this with nested for-loops:
for key, value_list in my_dict.iteritems():
for value in value_list:
output_file.write("{}\t{}\n".format(key, value))
this may also work...
#key = "one"
#value = ['1','2','3']
print '\n'.join(map(lambda i: key+'\t'+str(i), value))
I have two equal-length 1D numpy arrays, id and data, where id is a sequence of repeating, ordered integers that define sub-windows on data. For example:
id data
1 2
1 7
1 3
2 8
2 9
2 10
3 1
3 -10
I would like to aggregate data by grouping on id and taking either the max or the min.
In SQL, this would be a typical aggregation query like SELECT MAX(data) FROM tablename GROUP BY id ORDER BY id.
Is there a way I can avoid Python loops and do this in a vectorized manner?
I've been seeing some very similar questions on stack overflow the last few days. The following code is very similar to the implementation of numpy.unique and because it takes advantage of the underlying numpy machinery, it is most likely going to be faster than anything you can do in a python loop.
import numpy as np
def group_min(groups, data):
# sort with major key groups, minor key data
order = np.lexsort((data, groups))
groups = groups[order] # this is only needed if groups is unsorted
data = data[order]
# construct an index which marks borders between groups
index = np.empty(len(groups), 'bool')
index[0] = True
index[1:] = groups[1:] != groups[:-1]
return data[index]
#max is very similar
def group_max(groups, data):
order = np.lexsort((data, groups))
groups = groups[order] #this is only needed if groups is unsorted
data = data[order]
index = np.empty(len(groups), 'bool')
index[-1] = True
index[:-1] = groups[1:] != groups[:-1]
return data[index]
In pure Python:
from itertools import groupby, imap, izip
from operator import itemgetter as ig
print [max(imap(ig(1), g)) for k, g in groupby(izip(id, data), key=ig(0))]
# -> [7, 10, 1]
A variation:
print [data[id==i].max() for i, _ in groupby(id)]
# -> [7, 10, 1]
Based on #Bago's answer:
import numpy as np
# sort by `id` then by `data`
ndx = np.lexsort(keys=(data, id))
id, data = id[ndx], data[ndx]
# get max()
print data[np.r_[np.diff(id), True].astype(np.bool)]
# -> [ 7 10 1]
If pandas is installed:
from pandas import DataFrame
df = DataFrame(dict(id=id, data=data))
print df.groupby('id')['data'].max()
# id
# 1 7
# 2 10
# 3 1
I'm fairly new to Python and Numpy but, it seems like you can use the .at method of ufuncs rather than reduceat:
import numpy as np
data_id = np.array([0,0,0,1,1,1,1,2,2,2,3,3,3,4,5,5,5])
data_val = np.random.rand(len(data_id))
ans = np.empty(data_id[-1]+1) # might want to use max(data_id) and zeros instead
np.maximum.at(ans,data_id,data_val)
For example:
data_val = array([ 0.65753453, 0.84279716, 0.88189818, 0.18987882, 0.49800668,
0.29656994, 0.39542769, 0.43155428, 0.77982853, 0.44955868,
0.22080219, 0.4807312 , 0.9288989 , 0.10956681, 0.73215416,
0.33184318, 0.10936647])
ans = array([ 0.98969952, 0.84044947, 0.63460516, 0.92042078, 0.75738113,
0.37976055])
Of course this only makes sense if your data_id values are suitable for use as indices (i.e. non-negative integers and not huge...presumably if they are large/sparse you could initialize ans using np.unique(data_id) or something).
I should point out that the data_id doesn't actually need to be sorted.
with only numpy and without loops:
id = np.asarray([1,1,1,2,2,2,3,3])
data = np.asarray([2,7,3,8,9,10,1,-10])
# max
_ndx = np.argsort(id)
_id, _pos = np.unique(id[_ndx], return_index=True)
g_max = np.maximum.reduceat(data[_ndx], _pos)
# min
_ndx = np.argsort(id)
_id, _pos = np.unique(id[_ndx], return_index=True)
g_min = np.minimum.reduceat(data[_ndx], _pos)
# compare results with pandas groupby
np_group = pd.DataFrame({'min':g_min, 'max':g_max}, index=_id)
pd_group = pd.DataFrame({'id':id, 'data':data}).groupby('id').agg(['min','max'])
(pd_group.values == np_group.values).all() # TRUE
Ive packaged a version of my previous answer in the numpy_indexed package; its nice to have this all wrapped up and tested in a neat interface; plus it has a lot more functionality as well:
import numpy_indexed as npi
group_id, group_max_data = npi.group_by(id).max(data)
And so on
A slightly faster and more general answer than the already accepted one; like the answer by joeln it avoids the more expensive lexsort, and it works for arbitrary ufuncs. Moreover, it only demands that the keys are sortable, rather than being ints in a specific range. The accepted answer may still be faster though, considering the max/min isn't explicitly computed. The ability to ignore nans of the accepted solution is neat; but one may also simply assign nan values a dummy key.
import numpy as np
def group(key, value, operator=np.add):
"""
group the values by key
any ufunc operator can be supplied to perform the reduction (np.maximum, np.minimum, np.substract, and so on)
returns the unique keys, their corresponding per-key reduction over the operator, and the keycounts
"""
#upcast to numpy arrays
key = np.asarray(key)
value = np.asarray(value)
#first, sort by key
I = np.argsort(key)
key = key[I]
value = value[I]
#the slicing points of the bins to sum over
slices = np.concatenate(([0], np.where(key[:-1]!=key[1:])[0]+1))
#first entry of each bin is a unique key
unique_keys = key[slices]
#reduce over the slices specified by index
per_key_sum = operator.reduceat(value, slices)
#number of counts per key is the difference of our slice points. cap off with number of keys for last bin
key_count = np.diff(np.append(slices, len(key)))
return unique_keys, per_key_sum, key_count
names = ["a", "b", "b", "c", "d", "e", "e"]
values = [1.2, 4.5, 4.3, 2.0, 5.67, 8.08, 9.01]
unique_keys, reduced_values, key_count = group(names, values)
print 'per group mean'
print reduced_values / key_count
unique_keys, reduced_values, key_count = group(names, values, np.minimum)
print 'per group min'
print reduced_values
unique_keys, reduced_values, key_count = group(names, values, np.maximum)
print 'per group max'
print reduced_values
I think this accomplishes what you're looking for:
[max([val for idx,val in enumerate(data) if id[idx] == k]) for k in sorted(set(id))]
For the outer list comprehension, from right to left, set(id) groups the ids, sorted() sorts them, for k ... iterates over them, and max takes the max of, in this case, another list comprehension. So moving to that inner list comprehension: enumerate(data) returns both the index and value from data, if id[val] == k picks out the data members corresponding to id k.
This iterates over the full data list for each id. With some preprocessing into sublists, it might be possible to speed it up, but it won't be a one-liner then.
The following solution only requires a sort on the data (not a lexsort) and does not require finding boundaries between groups. It relies on the fact that if o is an array of indices into r then r[o] = x will fill r with the latest value x for each value of o, such that r[[0, 0]] = [1, 2] will return r[0] = 2. It requires that your groups are integers from 0 to number of groups - 1, as for numpy.bincount, and that there is a value for every group:
def group_min(groups, data):
n_groups = np.max(groups) + 1
result = np.empty(n_groups)
order = np.argsort(data)[::-1]
result[groups.take(order)] = data.take(order)
return result
def group_max(groups, data):
n_groups = np.max(groups) + 1
result = np.empty(n_groups)
order = np.argsort(data)
result[groups.take(order)] = data.take(order)
return result