I have a DataFrame with two pandas Series as follow:
value accepted_values
0 1 [1, 2, 3, 4]
1 2 [5, 6, 7, 8]
I would like to efficiently check if the value is in accepted_values using pandas methods.
I already know I can do something like the following, but I'm interested in a faster approach if there is one (took around 27 seconds on 1 million rows DataFrame)
import pandas as pd
df = pd.DataFrame({"value":[1, 2], "accepted_values": [[1,2,3,4], [5, 6, 7, 8]]})
def check_first_in_second(values: pd.Series):
return values[0] in values[1]
are_in_accepted_values = df[["value", "accepted_values"]].apply(
check_first_in_second, axis=1
)
if not are_in_accepted_values.all():
raise AssertionError("Not all value in accepted_values")
I think if create DataFrame with list column you can compare by DataFrame.eq and test if match at least one value per row by DataFrame.any:
df1 = pd.DataFrame(df["accepted_values"].tolist(), index=df.index)
are_in_accepted_values = df1.eq(df["value"]).any(axis=1).all()
Another idea:
are_in_accepted_values = all(v in a for v, a in df[["value", "accepted_values"]].to_numpy())
I found a little optimisation to your second idea. Using a bit more numpy than pandas makes it faster (more than 3x, tested with time.perf_counter()).
values = df["value"].values
accepted_values = df["accepted_values"].values
are_in_accepted_values = all(s in e for s, e in np.column_stack([values, accepted_values]))
Related
Given a dataframe, I am looking for rows where two out of three values are in common, regardless of the columns, hence order, in which they appear. I would like to then collect those common pairs.
Please note
a couple of values can appear at most in two rows
a value can appear only once in a row
I would like to know what the most efficient/elegant way is in numpy or pandas to solve this problem.
For example, taking as input the dataframe
d = {'col1': [1, 2,5,1], 'col2': [1, 7,1,2],'col3': [3, 3,1,7]}
df = pd.DataFrame(data=d)
col1 col2 col3
0 1 2 3
1 2 7 3
2 5 1 2
3 9 2 7
I expect as result an array, list, something as
1 2
2 3
2 7
as the values (1,2) , (2,3) and (2,7) are present in two rows (first and third, first and second, and second and forth respectively).
I cannot find a concise solution.
At the moment I skecthed a numpy solution such as
def func(x):
rows, columns = x.shape[0], x.shape[1]
res = []
for i in range(0,rows):
for j in range(i+1, rows):
aux = np.intersect1d(x[i,:], x[j,:])
if aux.size>1:
res.append(aux)
return res
which outputs
func(df.values)
Out: [array([2, 3]), array([1, 2]), array([2, 7])]
It looks well cumbersome, how could get it done with one of those cool numpy/pandas one-liners?
I would suggest using python built in set operations to do most of the heavy lifting, just apply them with pandas:
import itertools
import pandas as pd
d = {'col1': [1, 2,5,9], 'col2': [2, 7,1,2],'col3': [3, 3,2,7]}
df = pd.DataFrame(data=d)
pairs = df.apply(set, axis=1).apply(lambda x: set(itertools.combinations(x, 2))).explode()
out = set(pairs[pairs.duplicated()])
Output:
{(2, 3), (1, 2), (2, 7)}
Optionally to get it in list[np.ndarray] format:
out = list(map(np.array, out))
Similar approach to that of #Chrysophylaxs but in pure python:
from itertools import combinations
from collections import Counter
c = Counter(s for x in df.to_numpy().tolist() for s in set(combinations(set(x), r=2)))
out = [k for k,v in c.items() if v>1]
# [(2, 3), (1, 2), (2, 7)]
df=df.assign(col4=df.index)
def function1(ss:pd.Series):
ss1=ss.value_counts().loc[lambda ss:ss>=2]
return ss1.index.tolist() if ss1.size>=2 else None
df.merge(df,how='cross',suffixes=('','_2')).query("col4!=col4_2").filter(regex=r'col[^4]', axis=1)\
.apply(function1,axis=1).dropna().drop_duplicates()
out
1 [2, 3]
2 [1, 2]
7 [2, 7]
i have the following issue:
when i use .loc funtion it returns a series not a single value with no index.
As i need to do some math operation with the selected cells. the function that i am using is:
import pandas as pd
data = [[82,1], [30, 2], [3.7, 3]]
df = pd.DataFrame(data, columns = ['Ah-Step', 'State'])
df['Ah-Step'].loc[df['State']==2]+ df['Ah-Step'].loc[df['State']==3]
.values[0] will do what OP wants.
Assuming one wants to obtain the value 30, the following will do the work
df.loc[df['State'] == 2, 'Ah-Step'].values[0]
print(df)
[Out]: 30.0
So, in OP's specific case, the operation 30+3.7 could be done as follows
df.loc[df['State'] == 2, 'Ah-Step'].values[0] + df['Ah-Step'].loc[df['State']==3].values[0]
[Out]: 33.7
I have a pandas DataFrame, which contains 610 rows, and every row contains a nested list of coordinate pairs, it looks like that:
[1377778.4800000004, 6682395.377599999] is one coordinate pair.
I want to unnest every row, so instead of one row containing a list of coordinates I will have one row for every coordinate pair, i.e.:
I've tried s.apply(pd.Series).stack() from this question Split nested array values from Pandas Dataframe cell over multiple rows but unfortunately that didn't work.
Please any ideas? Many thanks in advance!
Here my new answer to your problem. I used "reduce" to flatten your nested array and then I used "itertools chain" to turn everything into a 1d list. After that I reshaped the list into a 2d array which allows you to convert it to the dataframe that you need. I tried to be as generic as possible. Please let me know if there are any problems.
#libraries
import operator
from functools import reduce
from itertools import chain
#flatten lists of lists using reduce. Then turn everything into a 1d list using
#itertools chain.
reduced_coordinates = list(chain.from_iterable(reduce(operator.concat,
geometry_list)))
#reshape the coordinates 1d list to a 2d and convert it to a dataframe
df = pd.DataFrame(np.reshape(reduced_coordinates, (-1, 2)))
df.columns = ['X', 'Y']
One thing you can do is use numpy. It allows you to perform a lot of list/ array operations in a fast and efficient way. This includes "unnesting" (reshaping) lists. Then you only have to convert to pandas dataframe.
For example,
import numpy as np
#your list
coordinate_list = [[[1377778.4800000004, 6682395.377599999],[6582395.377599999, 2577778.4800000004], [6582395.377599999, 2577778.4800000004]]]
#convert list to array
coordinate_array = numpy.array(coordinate_list)
#print shape of array
coordinate_array.shape
#reshape array into pairs of
reshaped_array = np.reshape(coordinate_array, (3, 2))
df = pd.DataFrame(reshaped_array)
df.columns = ['X', 'Y']
The output will look like this. Let me know if there is something I am missing.
import pandas as pd
import numpy as np
data = np.arange(500).reshape([250, 2])
cols = ['coord']
new_data = []
for item in data:
new_data.append([item])
df = pd.DataFrame(data=new_data, columns=cols)
print(df.head())
def expand(row):
row['x'] = row.coord[0]
row['y'] = row.coord[1]
return row
df = df.apply(expand, axis=1)
df.drop(columns='coord', inplace=True)
print(df.head())
RESULT
coord
0 [0, 1]
1 [2, 3]
2 [4, 5]
3 [6, 7]
4 [8, 9]
x y
0 0 1
1 2 3
2 4 5
3 6 7
4 8 9
I'm trying to save a named tuple n=NamedTuple(value1='x'=, value2='y') in a row of a pandas dataframe.
The problem is that the named tuple is showing a length of 2 because it has 2 parameters in my case (value1 and value2), so it doesn't fit it into a single cell of the dataframe.
How can I achieve that the named tuple is written into every call of a row of a dataframe?
df['columnd1']=n
an example:
from collections import namedtuple
import pandas as pd
n = namedtuple("test", ['param1', 'param2'])
n1 = n(param1='1', param2='2')
df = pd.DataFrame({"A": [1, 2, 3], "B": [4, 5, 6]})
df['nt'] = n1
print(df)
I don't really understand what you're trying to do, but if you want to put that named tuple in every row of a new column (i.e. like a scalar) then you can't rely on broadcasting but should instead replicate it yourself:
df['nt'] = [n1 for _ in range(df.shape[0])]
I want to visualize my data into box plots that are grouped by another variable shown here in my terrible drawing:
So what I do is to use a pandas series variable to tell pandas that I have grouped variables so this is what I do:
import pandas as pd
import seaborn as sns
#example data for reproduciblity
a = pd.DataFrame(
[
[2, 1],
[4, 2],
[5, 1],
[10, 2],
[9, 2],
[3, 1]
])
#converting second column to Series
a.ix[:,1] = pd.Series(a.ix[:,1])
#Plotting by seaborn
sns.boxplot(a, groupby=a.ix[:,1])
And this is what I get:
However, what I would have expected to get was to have two boxplots each describing only the first column, grouped by their corresponding column in the second column (the column converted to Series), while the above plot shows each column separately which is not what I want.
A column in a Dataframe is already a Series, so your conversion is not necessary. Furthermore, if you only want to use the first column for both boxplots, you should only pass that to Seaborn.
So:
#example data for reproduciblity
df = pd.DataFrame(
[
[2, 1],
[4, 2],
[5, 1],
[10, 2],
[9, 2],
[3, 1]
], columns=['a', 'b'])
#Plotting by seaborn
sns.boxplot(df.a, groupby=df.b)
I changed your example a little bit, giving columns a label makes it a bit more clear in my opinion.
edit:
If you want to plot all columns separately you (i think) basically want all combinations of the values in your groupby column and any other column. So if you Dataframe looks like this:
a b grouper
0 2 5 1
1 4 9 2
2 5 3 1
3 10 6 2
4 9 7 2
5 3 11 1
And you want boxplots for columns a and b while grouped by the column grouper. You should flatten the columns and change the groupby column to contain values like a1, a2, b1 etc.
Here is a crude way which i think should work, given the Dataframe shown above:
dfpiv = df.pivot(index=df.index, columns='grouper')
cols_flat = [dfpiv.columns.levels[0][i] + str(dfpiv.columns.levels[1][j]) for i, j in zip(dfpiv.columns.labels[0], dfpiv.columns.labels[1])]
dfpiv.columns = cols_flat
dfpiv = dfpiv.stack(0)
sns.boxplot(dfpiv, groupby=dfpiv.index.get_level_values(1))
Perhaps there are more fancy ways of restructuring the Dataframe. Especially the flattening of the hierarchy after pivoting is hard to read, i dont like it.
This is a new answer for an old question because in seaborn and pandas are some changes through version updates. Because of this changes the answer of Rutger is not working anymore.
The most important changes are from seaborn==v0.5.x to seaborn==v0.6.0. I quote the log:
Changes to boxplot() and violinplot() will probably be the most disruptive. Both functions maintain backwards-compatibility in terms of the kind of data they can accept, but the syntax has changed to be more similar to other seaborn functions. These functions are now invoked with x and/or y parameters that are either vectors of data or names of variables in a long-form DataFrame passed to the new data parameter.
Let's now go through the examples:
# preamble
import pandas as pd # version 1.1.4
import seaborn as sns # version 0.11.0
sns.set_theme()
Example 1: Simple Boxplot
df = pd.DataFrame([[2, 1] ,[4, 2],[5, 1],
[10, 2],[9, 2],[3, 1]
], columns=['a', 'b'])
#Plotting by seaborn with x and y as parameter
sns.boxplot(x='b', y='a', data=df)
Example 2: Boxplot with grouper
df = pd.DataFrame([[2, 5, 1], [4, 9, 2],[5, 3, 1],
[10, 6, 2],[9, 7, 2],[3, 11, 1]
], columns=['a', 'b', 'grouper'])
# usinge pandas melt
df_long = pd.melt(df, "grouper", var_name='a', value_name='b')
# join two columns together
df_long['a'] = df_long['a'].astype(str) + df_long['grouper'].astype(str)
sns.boxplot(x='a', y='b', data=df_long)
Example 3: rearanging the DataFrame to pass is directly to seaborn
def df_rename_by_group(data:pd.DataFrame, col:str)->pd.DataFrame:
'''This function takes a DataFrame, groups by one column and returns
a new DataFrame where the old columnnames are extended by the group item.
'''
grouper = df.groupby(col)
max_length_of_group = max([len(values) for item, values in grouper.indices.items()])
_df = pd.DataFrame(index=range(max_length_of_group))
for i in grouper.groups.keys():
helper = grouper.get_group(i).drop(col, axis=1).add_suffix(str(i))
helper.reset_index(drop=True, inplace=True)
_df = _df.join(helper)
return _df
df = pd.DataFrame([[2, 5, 1], [4, 9, 2],[5, 3, 1],
[10, 6, 2],[9, 7, 2],[3, 11, 1]
], columns=['a', 'b', 'grouper'])
df_new = df_rename_by_group(data=df, col='grouper')
sns.boxplot(data=df_new)
I really hope this answer helps to avoid some confusion.
sns.boxplot() doesnot take groupby.
Probably you are gonna see
TypeError: boxplot() got an unexpected keyword argument 'groupby'.
The best idea to group data and use in boxplot passing the data as groupby dataframe value.
import seaborn as sns
grouDataFrame = nameDataFrame(['A'])['B'].agg(sum).reset_index()
sns.boxplot(y='B', x='A', data=grouDataFrame)
Here B column data contains numeric value and grouped is done on the basis of A. All the grouped value with their respective column are added and boxplot diagram is plotted. Hope this helps.