I have a list of tuples that represent DataFrame index row number and a column name, in a form:
[(12, 'col3'), (16, 'col7'), ...].
I need to be able to find rows/column values that correspond to those tuple values in another dataframe and mark them red for example. Usually I use
df.style.apply(...)
from here: https://pandas.pydata.org/pandas-docs/stable/style.html and it works but in this case I am not sure how to map those tuple values with a dataframe in a function. Any help is much appreciated.
You can use custom function with at for set values by tups:
tups = [(12, 'col3'), (16, 'col7'), ...]
def highlight(x):
r = 'background-color: red'
df1 = pd.DataFrame('', index=x.index, columns=x.columns)
#rewrite values by selecting by tuples
for i, c in tups:
df1.at[i, c] = r
return df1
df.style.apply(highlight, axis=None)
Related
In Pyspark, I want to combine concat_ws and coalesce whilst using the list method. For example I know this works:
from pyspark.sql.functions import concat_ws, col
df = spark.createDataFrame([["A", "B"], ["C", None], [None, "D"]]).toDF("Type", "Segment")
#display(df)
df = df.withColumn("concat_ws2", concat_ws(':', coalesce('Type', lit("")), coalesce('Segment', lit(""))))
display(df)
But I want to be able to utilise the *[list] method so I don't have to list out all the columns within that bit of code, i.e. something like this instead:
from pyspark.sql.functions import concat_ws, col
df = spark.createDataFrame([["A", "B"], ["C", None], [None, "D"]]).toDF("Type", "Segment")
list = ["Type", "Segment"]
df = df.withColumn("almost_desired_output", concat_ws(':', *list))
display(df)
However as you can see, I want to be able to coalesce NULL with a blank, but not sure if that's possible using the *[list] method or do I really have to list out all the columns?
This would work:
Iterate over list of columns names
df=df.withColumn("almost_desired_output", concat_ws(':', *[coalesce(name, lit('')).alias(name) for name in df.schema.names]))
Output:
Or, Use fill - it'll fill all the null values across all columns of Dataframe (but this changes in the actual column, which may can break some use-cases)
df.na.fill("").withColumn("almost_desired_output", concat_ws(':', *list)
Or, Use selectExpr (again this changes in the actual column, which may can break some use-cases)
list = ["Type", "Segment"] # or just use df.schema.names
list2 = ["coalesce(type,' ') as Type", "coalesce(Segment,' ') as Segment"]
df=df.selectExpr(list2).withColumn("almost_desired_output", concat_ws(':', *list))
I am trying to set max decimal values upto 2 digit for result of a nested list. I have already tried to set precision and tried other things but can not find a way.
r_ij_matrix = variables[1]
print(type(r_ij_matrix))
print(type(r_ij_matrix[0]))
pd.set_option('display.expand_frame_repr', False)
pd.set_option("display.precision", 2)
data = pd.DataFrame(r_ij_matrix, columns= Attributes, index= Names)
df = data.style.set_table_styles([dict(selector='th', props=[('text-align', 'center')])])
df.set_properties(**{'text-align': 'center'})
df.set_caption('Table: Combined Decision Matrix')
You can solve your problem with the apply() method of the dataframe. You can do something like that :
df.apply(lambda x: [[round(elt, 2) for elt in list_] for list_ in x])
Solved it by copying the list to another with the desired decimal points. Thanks everyone.
rij_matrix = variables[1]
rij_nparray = np.empty([8, 6, 3])
for i in range(8):
for j in range(6):
for k in range(3):
rij_nparray[i][j][k] = round(rij_matrix[i][j][k], 2)
rij_list = rij_nparray.tolist()
pd.set_option('display.expand_frame_repr', False)
data = pd.DataFrame(rij_list, columns= Attributes, index= Names)
df = data.style.set_table_styles([dict(selector='th', props=[('text-align', 'center')])])
df.set_properties(**{'text-align': 'center'})
df.set_caption('Table: Normalized Fuzzy Decision Matrix (r_ij)')
applymap seems to be good here:
but there is a BUT: be aware that it is propably not the best idea to store lists as values of a df, you just give up the functionality of pandas. and also after formatting them like this, there are stored as strings. This (if really wanted) should only be for presentation.
df.applymap(lambda lst: list(map("{:.2f}".format, lst)))
Output:
A B
0 [2.05, 2.28, 2.49] [3.11, 3.27, 3.42]
1 [2.05, 2.28, 2.49] [3.11, 3.27, 3.42]
2 [2.05, 2.28, 2.49] [3.11, 3.27, 3.42]
Used Input:
df = pd.DataFrame({
'A': [[2.04939015319192, 2.280350850198276, 2.4899799195977463],
[2.04939015319192, 2.280350850198276, 2.4899799195977463],
[2.04939015319192, 2.280350850198276, 2.4899799195977463]],
'B': [[3.1144823004794873, 3.271085446759225, 3.420526275297414],
[3.1144823004794873, 3.271085446759225, 3.420526275297414],
[3.1144823004794873, 3.271085446759225, 3.420526275297414]]})
I have a data frame that looks like
d = {'col1': ['a,a,b', 'a,c,c,b'], 'col2': ['a,a,b', 'a,b,b,a']}
pd.DataFrame(data=d)
expected output
d={'col1':['a,b','a,c,b'],'col2':['a,b','a,b,a']}
I have tried like this :
arr = ['a', 'a', 'b', 'a', 'a', 'c','c']
print([x[0] for x in groupby(arr)])
How do I remove the duplicate entries in each row and column of dataframe?
a,a,b,c should be a,b,c
From what I understand, you don't want to include values which repeat in a sequence, you can try with this custom function:
def myfunc(x):
s=pd.Series(x.split(','))
res=s[s.ne(s.shift())]
return ','.join(res.values)
print(df.applymap(myfunc))
col1 col2
0 a,b a,b
1 a,c,b a,b,a
Another function can be created with itertools.groupby such as :
from itertools import groupby
def myfunc(x):
l=[x[0] for x in groupby(x.split(','))]
return ','.join(l)
You could define a function to help with this, then use .applymap to apply it to all columns (or .apply one column at a time):
d = {'col1': ['a,a,b', 'a,c,c,b'], 'col2': ['a,a,b', 'a,b,b,a']}
df = pd.DataFrame(data=d)
def remove_dups(string):
split = string.split(',') # split string into a list
uniques = set(split) # remove duplicate list elements
return ','.join(uniques) # rejoin the list elements into a string
result = df.applymap(remove_dups)
This returns:
col1 col2
0 a,b a,b
1 a,c,b a,b
Edit: This looks slightly different to your expected output, why do you expect a,b,a for the second row in col2?
Edit2: to preserve the original order, you can replace the set() function with unique_everseen()
from more_itertools import unique_everseen
.
.
.
uniques = unique_everseen(split)
I have a dictionary:
d = {'A-A': 1, 'A-B':2, 'A-C':3, 'B-A':5, 'B-B':1, 'B-C':5, 'C-A':3,
'C-B':4, 'C-C': 9}
and a list:
L = [A,B,C]
I have a DataFrame:
df =pd.DataFrame(columns = L, index=L)
I would like to fill each row in df by values in dictionary based on dictionary keys.For example:
A B C
A 1 2 3
B 5 1 5
C 3 4 9
I tried doing that by:
df.loc[L[0]]=[1,2,3]
df.loc[L[1]]=[5,1,5]
df.loc[L[2]] =[3,4,9]
Is there another way to do that especially when there is a huge data?
Thank you for help
Here is another way that I can think of:
import numpy as np
import pandas as pd
# given
d = {'A-A': 1, 'A-B':2, 'A-C':3, 'B-A':5, 'B-B':1, 'B-C':5, 'C-A':3,
'C-B':4, 'C-C': 9}
L = ['A', 'B', 'C']
# copy the key values into a numpy array
z = np.asarray(list(d.values()))
# reshape the array according to your DataFrame
z_new = np.reshape(z, (3, 3))
# copy it into your DataFrame
df = pd.DataFrame(z_new, columns = L, index=L)
This should do the trick, though it's probably not the best way:
for index in L:
prefix = index + "-"
df.loc[index] = [d.get(prefix + column, 0) for column in L]
Calculating the prefix separately beforehand is probably slower for a small list and probably faster for a large list.
Explanation
for index in L:
This iterates through all of the row names.
prefix = index + "-"
All of the keys for each row start with index + "-", e.g. "A-", "B-"… etc..
df.loc[index] =
Set the contents of the entire row.
[ for column in L]
The same as your comma thing ([1, 2, 3]) just for an arbitrary number of items. This is called a "list comprehension".
d.get( , 0)
This is the same as d[ ] but returns 0 if it can't find anything.
prefix + column
Sticks the column on the end, e.g. "A-" gives "A-A", "A-B"…
I have two dataframes, both indexed with timestamps. I would like to preserve the order of the columns in the first dataframe that is merged.
For example:
#required packages
import pandas as pd
import numpy as np
# defining stuff
num_periods_1 = 11
num_periods_2 = 4
# create sample time series
dates1 = pd.date_range('1/1/2000 00:00:00', periods=num_periods_1, freq='10min')
dates2 = pd.date_range('1/1/2000 01:30:00', periods=num_periods_2, freq='10min')
column_names_1 = ['C', 'B', 'A']
column_names_2 = ['B', 'C', 'D']
df1 = pd.DataFrame(np.random.randn(num_periods_1, len(column_names_1)), index=dates1, columns=column_names_1)
df2 = pd.DataFrame(np.random.randn(num_periods_2, len(column_names_2)), index=dates2, columns=column_names_2)
df3 = df1.merge(df2, how='outer', left_index=True, right_index=True, suffixes=['_1', '_2'])
print("\nData Frame Three:\n", df3)
The above code generates two data frames the first with columns C, B, and A. The second dataframe has columns B, C, and D. The current output has the columns in the following order; C_1, B_1, A, B_2, C_2, D. What I want the columns from the output of the merge to be C_1, C_2, B_1, B_2, A_1, D_2. The order of the columns is preserved from the first data frame and any data similar to the second data frame is added next to the corresponding data.
Could there be a setting in merge or can I use sort_index to do this?
EDIT: Maybe a better way to phrase the sorting process would be to call it uncollated. Where each column is put together and so on.
Using an OrderedDict, as you suggested.
from collections import OrderedDict
from itertools import chain
c = df3.columns.tolist()
o = OrderedDict()
for x in c:
o.setdefault(x.split('_')[0], []).append(x)
c = list(chain.from_iterable(o.values()))
df3 = df3[c]
An alternative that involves extracting the prefixes and then calling sorted on the index.
# https://stackoverflow.com/a/46839182/4909087
p = [s[0] for s in c]
c = sorted(c, key=lambda x: (p.index(x[0]), x))
df = df[c]