My table looks like this(df):
category
product_in_cat
cat1
[A,B,C]
cat2
[E,F,G]
"category" is str, and product_in_cat is list type. I have a list:product=[A,B,G]
I want to get a final [dict(str:list)] looks like:
[{cat1:[A,B]},{cat2:[G]}]
I think I can use below code:
list1=[]
for inde,row in df.iterrows():
list1.append.({row['category']:row['product_in_cat'] in product})
I know this part is not correct,row['product_in_cat'] in product but I am not sure how to filter out the list column base on the given "product" list. Please help, and thank you in advance!
You can use np.intersect1d to find the common part of two lists:
import numpy as np
df_ = df['product_in_cat'].apply(lambda x: np.intersect1d(x, product).tolist())
l = [{k: v} for k, v in zip(df['category'], df_)]
print(l)
[{'cat1': ['A', 'B']}, {'cat2': ['G']}]
You can use convert each list in the column to a set and use intersection with the external product list:
import pandas as pd
lst = ['A','B','G']
data = {'category':['cat 1','cat 2'],
'product_in_cat': [ ['A','B','C'] ,['E','F','G']]}
df = pd.DataFrame(data)
dict(zip(df['category'],df['product_in_cat'].apply(lambda x: set(x).intersection(lst))))
#output
{'cat 1': {'A', 'B'}, 'cat 2': {'G'}}
Related
is there a possibility to adjust the strings according to the order for example 1.wav, 2.wav 3.wav etc. and the ID accordingly with ID: 1, 2, 3 etc?
i have already tried several sorting options do any of you have any ideas?
Thank you in advance
dataframe output
def createSampleDF(audioPath):
data = []
for file in Path(audioPath).glob('*.wav'):
print(file)
data.append([os.path.basename(file), file])
df_dataSet = pd.DataFrame(data, columns= ['audio_name', 'filePath'])
df_dataSet['ID'] = df_dataSet.index+1
df_dataSet = df_dataSet[['ID','audio_name','filePath']]
df_dataSet.sort_values(by=['audio_name'],inplace=True)
return df_dataSet
def createSamples(myAudioPath,savePath, sampleLength, overlap = 0):
cutSamples(myAudioPath=myAudioPath,savePath=savePath,sampleLength=sampleLength)
df_dataSet=createSampleDF(audioPath=savePath)
return df_dataSet
You can split the string, make it an integer, and then sort on multiple columns. See the pandas.Dataframe.sort_values for more info. If your links are more complicated you may need to design a regex to pull out the integers you want to sort on using pandas.Series.str.extract.
df = pd.DataFrame({
'ID':[1,2,3,4, 5],
'audio_name' : ['1.wav','10.wav','96.wav','3.wav','55.wav']})
(df
.assign(audio_name=lambda df_ : df_.audio_name.str.split('.', expand=True).iloc[:,0].astype('int'))
.sort_values(by=['audio_name','ID']))
I have a pandas dataframe with one column that contains an empty list in each cell.
I need to duplicate the dataframe, and append it at the bottom of the original dataframe, but with additional information in the list.
Here is a minimal code example:
df_main = pd.DataFrame([['a', []], ['b', []]], columns=['letter', 'mylist'])
> df_main
letter mylist
0 a []
1 b []
df_copy = df_main.copy()
for index, row in df_copy.iterrows():
row.mylist = row.mylist.append(1)
pd.concat([ df_copy,df_main], ignore_index=True)
> result:
letter mylist
0 a None
1 b None
2 a [1]
3 b [1]
As you can see there is a problem that the [] empty list was replaced by a None
Just to make sure, this is what I would like to have:
letter mylist
0 a []
1 b []
2 a [1]
3 b [1]
How can I achieve that?
append method on list return a None value, that's why None appears in the final dataframe. You may have use + operator for reassignment like this:
import pandas as pd
df_main = pd.DataFrame([['a', []], ['b', []]], columns=['letter', 'mylist'])
df_copy = df_main.copy()
for index, row in df_copy.iterrows():
row.mylist = row.mylist + list([1])
pd.concat([df_main, df_copy], ignore_index=True).head()
Output of this block of code:
letter mylist
0 a []
1 b []
2 a [1]
3 b [1]
A workaround to solve your problem would be to create a temporary column mylist2 with np.empty((len(df), 0)).tolist()) and use np.where() to change the None values of mylist to an empty list and then drop the empty column.
import pandas as pd, numpy as np
df_main = pd.DataFrame([['a', []], ['b', []]], columns=['letter', 'mylist'])
df_copy = df_main.copy()
for index, row in df_copy.iterrows():
row.mylist = row.mylist.append(1)
df = (pd.concat([df_copy,df_main], ignore_index=True)
.assign(mylist2=np.empty((len(df), 0)).tolist()))
df['mylist'] = np.where((df['mylist'].isnull()), df['mylist2'], df['mylist'])
df= df.drop('mylist2', axis=1)
df
Out[1]:
letter mylist
0 a []
1 b []
2 a [1]
3 b [1]
Not only does append method on list return a None value as indicated in the first answer, but both df_main and df_copy contain pointers to the same lists. So after:
for index, row in df_copy.iterrows():
row.mylist.append(1)
both dataframes have updated lists with one element. For your code to work as expected you can create a new list after you copy the dataframe:
df_copy = df_main.copy()
for index, row in df_copy.iterrows():
row.mylist = []
This question is another great example why we should not put objects in a dataframe.
I have a problem with the following scenario using PySpark version 2.0, I have a DataFrame with a column contains an array with start and end value, e.g.
[1000, 1010]
I would like to know how to create and compute another column which contains an array that holds all the values for the given range? the result of the generated range values column will be:
+--------------+-------------+-----------------------------+
| Description| Accounts| Range|
+--------------+-------------+-----------------------------+
| Range 1| [101, 105]| [101, 102, 103, 104, 105]|
| Range 2| [200, 203]| [200, 201, 202, 203]|
+--------------+-------------+-----------------------------+
Try this.
define the udf
def range_value(a):
start = a[0]
end = a[1] +1
return list(range(start,end))
from pyspark.sql import functions as F
from pyspark.sql import types as pt
df = spark.createDataFrame([("Range 1", list([101,105])), ("Range 2", list([200, 203]))],("Description", "Accounts"))
range_value= F.udf(range_value, pt.ArrayType(pt.IntegerType()))
df = df.withColumn('Range', range_value(F.col('Accounts')))
Output
you should use UDF (UDF sample)
Consider your pyspark data frame name is df, your data frame could be like this:
df = spark.createDataFrame(
[("Range 1", list([101,105])),
("Range 2", list([200, 203]))],
("Description", "Accounts"))
And your solution is like this:
import pyspark.sql.functions as F
import numpy as np
def make_range_number(arr):
number_range = np.arange(arr[0], arr[1]+1, 1).tolist()
return number_range
range_udf = F.udf(make_range_number)
df = df.withColumn("Range", range_udf(F.col("Accounts")))
Have a fun time!:)
df.groupby('columns').apply(''.join()), join all the cells to a string.
This is for a junior dataprocessor. In the past, I've tried many ways.
import pandas as pd
data = {'key':['a','b','c','a','b','c','a'], 'profit':
[12,3,4,5,6,7,9],'income':['j','d','d','g','d','t','d']}
df = pd.DataFrame(data)
df = df.set_index(‘key’)
#df2 is expected result
data2 = {'a':['12j5g9d'],'b':['3d6d'],'c':['4d7t']}
df2 = pd.DataFrame(data2)
df2 = df2.set_index(‘key’)
Here's a simple solution, where we first translate the integers to strings and then concatenate profit and income, then finally we concatenate all strings under the same key:
data = {'key':['a','b','c','a','b','c','a'], 'profit':
[12,3,4,5,6,7,9],'income':['j','d','d','g','d','t','d']}
df = pd.DataFrame(data)
df['profit_income'] = df['profit'].apply(str) + df['income']
res = df.groupby('key')['profit_income'].agg(''.join)
print(res)
output:
key
a 12j5g9d
b 3d6d
c 4d7t
Name: profit_income, dtype: object
This question can be solved couple different ways:
First add an extra column by concatenating the profit and income columns.
import pandas as pd
data = {'key':['a','b','c','a','b','c','a'], 'profit':
[12,3,4,5,6,7,9],'income':['j','d','d','g','d','t','d']}
df = pd.DataFrame(data)
df = df.set_index('key')
df['profinc']=df['profit'].astype(str)+df['income']
1) Using sum
df2=df.groupby('key').profinc.sum()
2) Using apply and join
df2=df.groupby('key').profinc.apply(''.join)
Results from both of the above would be the same:
key
a 12j5g9d
b 3d6d
c 4d7t
I have a dataframe with column names, and I want to find the one that contains a certain string, but does not exactly match it. I'm searching for 'spike' in column names like 'spike-2', 'hey spike', 'spiked-in' (the 'spike' part is always continuous).
I want the column name to be returned as a string or a variable, so I access the column later with df['name'] or df[name] as normal. I've tried to find ways to do this, to no avail. Any tips?
Just iterate over DataFrame.columns, now this is an example in which you will end up with a list of column names that match:
import pandas as pd
data = {'spike-2': [1,2,3], 'hey spke': [4,5,6], 'spiked-in': [7,8,9], 'no': [10,11,12]}
df = pd.DataFrame(data)
spike_cols = [col for col in df.columns if 'spike' in col]
print(list(df.columns))
print(spike_cols)
Output:
['hey spke', 'no', 'spike-2', 'spiked-in']
['spike-2', 'spiked-in']
Explanation:
df.columns returns a list of column names
[col for col in df.columns if 'spike' in col] iterates over the list df.columns with the variable col and adds it to the resulting list if col contains 'spike'. This syntax is list comprehension.
If you only want the resulting data set with the columns that match you can do this:
df2 = df.filter(regex='spike')
print(df2)
Output:
spike-2 spiked-in
0 1 7
1 2 8
2 3 9
This answer uses the DataFrame.filter method to do this without list comprehension:
import pandas as pd
data = {'spike-2': [1,2,3], 'hey spke': [4,5,6]}
df = pd.DataFrame(data)
print(df.filter(like='spike').columns)
Will output just 'spike-2'. You can also use regex, as some people suggested in comments above:
print(df.filter(regex='spike|spke').columns)
Will output both columns: ['spike-2', 'hey spke']
You can also use df.columns[df.columns.str.contains(pat = 'spike')]
data = {'spike-2': [1,2,3], 'hey spke': [4,5,6], 'spiked-in': [7,8,9], 'no': [10,11,12]}
df = pd.DataFrame(data)
colNames = df.columns[df.columns.str.contains(pat = 'spike')]
print(colNames)
This will output the column names: 'spike-2', 'spiked-in'
More about pandas.Series.str.contains.
# select columns containing 'spike'
df.filter(like='spike', axis=1)
You can also select by name, regular expression. Refer to: pandas.DataFrame.filter
df.loc[:,df.columns.str.contains("spike")]
Another solution that returns a subset of the df with the desired columns:
df[df.columns[df.columns.str.contains("spike|spke")]]
You also can use this code:
spike_cols =[x for x in df.columns[df.columns.str.contains('spike')]]
Getting name and subsetting based on Start, Contains, and Ends:
# from: https://stackoverflow.com/questions/21285380/find-column-whose-name-contains-a-specific-string
# from: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.str.contains.html
# from: https://cmdlinetips.com/2019/04/how-to-select-columns-using-prefix-suffix-of-column-names-in-pandas/
# from: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.filter.html
import pandas as pd
data = {'spike_starts': [1,2,3], 'ends_spike_starts': [4,5,6], 'ends_spike': [7,8,9], 'not': [10,11,12]}
df = pd.DataFrame(data)
print("\n")
print("----------------------------------------")
colNames_contains = df.columns[df.columns.str.contains(pat = 'spike')].tolist()
print("Contains")
print(colNames_contains)
print("\n")
print("----------------------------------------")
colNames_starts = df.columns[df.columns.str.contains(pat = '^spike')].tolist()
print("Starts")
print(colNames_starts)
print("\n")
print("----------------------------------------")
colNames_ends = df.columns[df.columns.str.contains(pat = 'spike$')].tolist()
print("Ends")
print(colNames_ends)
print("\n")
print("----------------------------------------")
df_subset_start = df.filter(regex='^spike',axis=1)
print("Starts")
print(df_subset_start)
print("\n")
print("----------------------------------------")
df_subset_contains = df.filter(regex='spike',axis=1)
print("Contains")
print(df_subset_contains)
print("\n")
print("----------------------------------------")
df_subset_ends = df.filter(regex='spike$',axis=1)
print("Ends")
print(df_subset_ends)