I want to find the count for the number of previous rows that have the a greater value than the current row in a column and store it in a new column. It would be like a rolling countif that goes back to the beginning of the column. The desired example output below shows the value column given and the count column I want to create.
Desired Output:
Value Count
5 0
7 0
4 2
12 0
3 4
4 3
1 6
I plan on using this code with a large dataframe so the fastest way possible is appreciated.
We can do subtract.outer from numpy , then get lower tri and find the value is less than 0, and sum the value per row
a = np.sum(np.tril(np.subtract.outer(df.Value.values,df.Value.values), k=0)<0, axis=1)
# results in array([0, 0, 2, 0, 4, 3, 6])
df['Count'] = a
IMPORTANT: this only works with pandas < 1.0.0 and the error seems to be a pandas bug. An issue is already created at https://github.com/pandas-dev/pandas/issues/35203
We can do this with expanding and applying a function which checks for values that are higher than the last element in the expanding array.
import pandas as pd
import numpy as np
# setup
df = pd.DataFrame([5,7,4,12,3,4,1], columns=['Value'])
# calculate countif
df['Count'] = df.Value.expanding(1).apply(lambda x: np.sum(np.where(x > x[-1], 1, 0))).astype('int')
Input
Value
0 5
1 7
2 4
3 12
4 3
5 4
6 1
Output
Value Count
0 5 0
1 7 0
2 4 2
3 12 0
4 3 4
5 4 3
6 1 6
count = []
for i in range(len(values)):
count = 0
for j in values[:i]:
if values[i] < j:
count += 1
count.append(count)
The below generator will do what you need. You may be able to further optimize this if needed.
def generator (data) :
i=0
count_dict ={}
while i<len(data) :
m=max(data)
v=data[i]
count_dict[v] =count_dict[v] +1 if v in count_dict else 1
t=sum([(count_dict[j] if j in count_dict else 0) for j in range(v+1,m)])
i +=1
yield t
d=[1, 5,7,3,5,8]
foo=generator (d)
result =[b for b in foo]
print(result)
Related
I have the following working code that sets 1 to "new_col" at the locations pointed by intervals dictated by starts and ends.
import pandas as pd
import numpy as np
df = pd.DataFrame({"a": np.arange(10)})
starts = [1, 5, 8]
ends = [1, 6, 10]
value = 1
df["new_col"] = 0
for s, e in zip(starts, ends):
df.loc[s:e, "new_col"] = value
print(df)
a new_col
0 0 0
1 1 1
2 2 0
3 3 0
4 4 0
5 5 1
6 6 1
7 7 0
8 8 1
9 9 1
I want these intervals to come from another dataframe pointer_df.
How to vectorize this?
pointer_df = pd.DataFrame({"starts": starts, "ends": ends})
Attempt:
df.loc[pointer_df["starts"]:pointer_df["ends"], "new_col"] = 2
print(df)
obviously doesn't work and gives
raise AssertionError("Start slice bound is non-scalar")
AssertionError: Start slice bound is non-scalar
EDIT:
it seems all answers use some kind of pythonic for loop.
the question was how to vectorize the operation above?
Is this not doable without for loops/list comprehentions?
You could do:
pointer_df = pd.DataFrame({"starts": starts, "ends": ends})
rang = np.arange(len(df))
indices = [i for s, e in pointer_df.to_numpy() for i in rang[slice(s, e + 1, None)]]
df.loc[indices, 'new_col'] = value
print(df)
Output
a new_col
0 0 0
1 1 1
2 2 0
3 3 0
4 4 0
5 5 1
6 6 1
7 7 0
8 8 1
9 9 1
If you want a method that do not uses uses any for loop or list comprehension, only relies on numpy, you could do:
def indices(start, end, ma=10):
limits = end + 1
lens = np.where(limits < ma, limits, end) - start
np.cumsum(lens, out=lens)
i = np.ones(lens[-1], dtype=int)
i[0] = start[0]
i[lens[:-1]] += start[1:]
i[lens[:-1]] -= limits[:-1]
np.cumsum(i, out=i)
return i
pointer_df = pd.DataFrame({"starts": starts, "ends": ends})
df.loc[indices(pointer_df.starts.values, pointer_df.ends.values, ma=len(df)), "new_col"] = value
print(df)
I adapted the method to your use case from the one in this answer.
for i,j in zip(pointer_df["starts"],pointer_df["ends"]):
print (i,j)
Apply same method but on your dictionary
I have a df like so:
Period Count
1 1
2 0
3 1
4 1
5 0
6 0
7 1
8 1
9 1
10 0
and I want to return a 'Event ID' in a new column if there are two or more consecutive occurrences of 1 in Count and a 0 if there is not. So in the new column each row would get a 1 based on this criteria being met in the column Count. My desired output would then be:
Period Count Event_ID
1 1 0
2 0 0
3 1 1
4 1 1
5 0 0
6 0 0
7 1 2
8 1 2
9 1 2
10 0 0
I have researched and found solutions that allow me to flag out consecutive group of similar numbers (e.g 1) but I haven't come across what I need yet. I would like to be able to use this method to count any number of consecutive occurrences, not just 2 as well. For example, sometimes I need to count 10 consecutive occurrences, I just use 2 in the example here.
This will do the job:
ones = df.groupby('Count').groups[1].tolist()
# creates a list of the indices with a '1': [0, 2, 3, 6, 7, 8]
event_id = [0] * len(df.index)
# creates a list of length 10 for Event_ID with all '0'
# find consecutive numbers in the list of ones (yields [2,3] and [6,7,8]):
for k, g in itertools.groupby(enumerate(ones), lambda ix : ix[0] - ix[1]):
sublist = list(map(operator.itemgetter(1), g))
if len(sublist) > 1:
for i in sublist:
event_id[i] = len(sublist)-1
# event_id is now [0, 0, 1, 1, 0, 0, 2, 2, 2, 0]
df['Event_ID'] = event_id
The for loop is adapted from this example (using itertools, other approaches are possible too).
I have following python data frame
data={'1':[1,1,1,1],'2':[1,1,1,1],'3':[1,1,1,1]}
df=pd.DataFrame(data)
I need to get the sum of the rows in a such away that my final output should be like this,
So in this desired output, the second column should contain the row sum up to second column of the original data frame. So on.
To get this output, I wrote the following code,
sum_mat=np.zeros(shape=(3,3))
numOfIteration=3
itr=list(range(0,numOfIteration))
for i in range(0,3):
for j in range(0,3):
while i <= itr[i]:
sum_mat[i,j]+= df.iloc[i,j]
print (sum_mat)
I am not getting an output here because the code is running forever (may be an infinite loop).
Can anyone suggest anything to get the desired output ?
May be there is more effective and easier way to do the same thing.
Thank you
UPDATE:
i update the for loop as follows,
for i in range(0,3):
for j in range(0,3):
while i <= itr[i]:
sum_mat[i,j] = df.iloc[:,0:i].sum(axis=1)
but it gives following error,
sum_mat[i,j] = df.iloc[:,0:i].sum(axis=1)
ValueError: setting an array element with a sequence.
this could work also
for i,row in df.iterrows(): #go through each row
df.loc[i]=df.loc[i].cumsum() #assign each row as the cumulative sum of the row
output:
>>> df
1 2 3
0 1 2 3
1 1 2 3
2 1 2 3
3 1 2 3
EDIT
can just do :
df=df.cumsum(axis=1)
sum_mat=np.zeros(shape=(3,3))
numOfIteration=3
itr=list(range(0,numOfIteration))
for i in range(0,3):
for j in range(0,3):
if j==0:
sum_mat[i,0]=df.iloc[i,0]
else:
sum_mat[i,j]=df.iloc[i,j]+sum_mat[i,j-1]
print (sum_mat)
This should work
Use cumsum() function to find the cumulative sum of the values seen so far along the column axis.
Ex.
import pandas as pd
data = {'1': [1, 1, 1, 1], '2': [1, 1, 1, 1], '3': [1, 1, 1, 1]}
df = pd.DataFrame(data)
print("before")
print(df)
df = df.cumsum(axis=1)
print("after")
print(df)
O/P:
before
1 2 3
0 1 1 1
1 1 1 1
2 1 1 1
3 1 1 1
after
1 2 3
0 1 2 3
1 1 2 3
2 1 2 3
3 1 2 3
I was wondering if anyone could help me with how to make a bar chart to show the frequencies of values in a Pandas Series.
I start with a Pandas DataFrame of shape (2000, 7), and from there I extract the last column. The column is shape (2000,).
The entries in the Series that I mentioned vary from 0 to 17, each with different frequencies, and I tried to plot them using a bar chart but faced some difficulties. Here is my code:
# First, I counted the number of occurrences.
count = np.zeros(max(data_val))
for i in range(count.shape[0]):
for j in range(data_val.shape[0]):
if (i == data_val[j]):
count[i] = count[i] + 1
'''
This gives us
count = array([192., 105., ... 19.])
'''
temp = np.arange(0, 18, 1) # Array for the x-axis.
plt.bar(temp, count)
I am getting an error on the last line of code, saying that the objects cannot be broadcast to a single shape.
What I ultimately want is a bar chart where each bar corresponds to an integer value from 0 to 17, and the height of each bar (i.e. the y-axis) represents the frequencies.
Thank you.
UPDATE
I decided to post the fixed code using the suggestions that people were kind enough to give below, just in case anybody facing similar issues will be able to see my revised code in the future.
data = pd.read_csv("./data/train.csv") # Original data is a (2000, 7) DataFrame
# data contains 6 feature columns and 1 target column.
# Separate the design matrix from the target labels.
X = data.iloc[:, :-1]
y = data['target']
'''
The next line of code uses pandas.Series.value_counts() on y in order to count
the number of occurrences for each label, and then proceeds to sort these according to
index (i.e. label).
You can also use pandas.DataFrame.sort_values() instead if you're interested in sorting
according to the number of frequencies rather than labels.
'''
y.value_counts().sort_index().plot.bar(x='Target Value', y='Number of Occurrences')
There was no need to use for loops if we use the methods that are built into the Pandas library.
The specific methods that were mentioned in the answers are pandas.Series.values_count(), pandas.DataFrame.sort_index(), and pandas.DataFrame.plot.bar().
I believe you need value_counts with Series.plot.bar:
df = pd.DataFrame({
'a':[4,5,4,5,5,4],
'b':[7,8,9,4,2,3],
'c':[1,3,5,7,1,0],
'd':[1,1,6,1,6,5],
})
print (df)
a b c d
0 4 7 1 1
1 5 8 3 1
2 4 9 5 6
3 5 4 7 1
4 5 2 1 6
5 4 3 0 5
df['d'].value_counts(sort=False).plot.bar()
If possible some value missing and need set it to 0 add reindex:
df['d'].value_counts(sort=False).reindex(np.arange(18), fill_value=0).plot.bar()
Detail:
print (df['d'].value_counts(sort=False))
1 3
5 1
6 2
Name: d, dtype: int64
print (df['d'].value_counts(sort=False).reindex(np.arange(18), fill_value=0))
0 0
1 3
2 0
3 0
4 0
5 1
6 2
7 0
8 0
9 0
10 0
11 0
12 0
13 0
14 0
15 0
16 0
17 0
Name: d, dtype: int64
Here's an approach using Seaborn
import numpy as np
import pandas as pd
import seaborn as sns
s = pd.Series(np.random.choice(17, 10))
s
# 0 10
# 1 13
# 2 12
# 3 0
# 4 0
# 5 5
# 6 13
# 7 9
# 8 11
# 9 0
# dtype: int64
val, cnt = np.unique(s, return_counts=True)
val, cnt
# (array([ 0, 5, 9, 10, 11, 12, 13]), array([3, 1, 1, 1, 1, 1, 2]))
sns.barplot(val, cnt)
Having a DataFrame with the following column:
df['A'] = [1,1,1,0,1,1,1,1,0,1]
What would be the best vectorized way to control the length of "1"-series by some limiting value? Let's say the limit is 2, then the resulting column 'B' must look like:
A B
0 1 1
1 1 1
2 1 0
3 0 0
4 1 1
5 1 1
6 1 0
7 1 0
8 0 0
9 1 1
One fully-vectorized solution is to use the shift-groupby-cumsum-cumcount combination1 to indicate where consecutive runs are shorter than 2 (or whatever limiting value you like). Then, & this new boolean Series with the original column:
df['B'] = ((df.groupby((df.A != df.A.shift()).cumsum()).cumcount() <= 1) & df.A)\
.astype(int) # cast the boolean Series back to integers
This produces the new column in the DataFrame:
A B
0 1 1
1 1 1
2 1 0
3 0 0
4 1 1
5 1 1
6 1 0
7 1 0
8 0 0
9 1 1
1 See the pandas cookbook; the section on grouping, "Grouping like Python’s itertools.groupby"
Another way (checking if previous two are 1):
In [443]: df = pd.DataFrame({'A': [1,1,1,0,1,1,1,1,0,1]})
In [444]: limit = 2
In [445]: df['B'] = map(lambda x: df['A'][x] if x < limit else int(not all(y == 1 for y in df['A'][x - limit:x])), range(len(df)))
In [446]: df
Out[446]:
A B
0 1 1
1 1 1
2 1 0
3 0 0
4 1 1
5 1 1
6 1 0
7 1 0
8 0 0
9 1 1
If you know that the values in the series will all be either 0 or 1, I think you can use a little trick involving convolution. Make a copy of your column (which need not be a Pandas object, it can just be a normal Numpy array)
a = df['A'].as_matrix()
and convolve it with a sequence of 1's that is one longer than the cutoff you want, then chop off the last cutoff elements. E.g. for a cutoff of 2, you would do
long_run_count = numpy.convolve(a, [1, 1, 1])[:-2]
The resulting array, in this case, gives the number of 1's that occur in the 3 elements prior to and including that element. If that number is 3, then you are in a run that has exceeded length 2. So just set those elements to zero.
a[long_run_count > 2] = 0
You can now assign the resulting array to a new column in your DataFrame.
df['B'] = a
To turn this into a more general method:
def trim_runs(array, cutoff):
a = numpy.asarray(array)
a[numpy.convolve(a, numpy.ones(cutoff + 1))[:-cutoff] > cutoff] = 0
return a