Hi, given this dataframe is it possible to fetch the Number value associated with certain conditions using df.loc? This is what i came up with so far.
if df.loc[(df["Tags"]=="Brunei") & (df["Type"]=="Host"),"Number"]:
I want the output to be 1. Is this the correct way to do it?
You're in the right way, but you have to pass ".values[0]" in the end of the .loc statement to extract the only value that you got in the pandas Series.
df = pd.DataFrame({
'Tags': ['Brunei', 'China'],
'Type': ['Host', 'Address'],
'Number': [1, 1192]
}
)
display(df)
series = df.loc[(df["Tags"]=="Brunei") & (df["Type"]=="Host"),"Number"]
print(type(series))
value = df.loc[(df["Tags"]=="Brunei") & (df["Type"]=="Host"),"Number"].values[0]
print(type(value))
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 series of web addresses, which I want to split them by the first '.'. For example, return 'google', if the web address is 'google.co.uk'
d1 = {'id':['1', '2', '3'], 'website':['google.co.uk', 'google.com.au', 'google.com']}
df1 = pd.DataFrame(data=d1)
d2 = {'id':['4', '5', '6'], 'website':['google.co.jp', 'google.com.tw', 'google.kr']}
df2 = pd.DataFrame(data=d2)
df_list = [df1, df2]
I use enumerate to iterate the dataframe list
for i, df in enumerate(df_list):
df_list[i]['website_segments'] = df['website'].str.split('.', n=1, expand=True)
Received error: ValueError: Wrong number of items passed 2, placement implies 1
You are splitting the website which gives you a list-like data structure. Think [google, co.uk]. You just want the first element of that list so:
for i, df in enumerate(df_list):
df_list[i]['website_segments'] = df['website'].str.split('.', n=1, expand=True)[0]
Another alternative is to use extract. It is also ~40% faster for your data:
for i, df in enumerate(df_list):
df_list[i]['website_segments'] = df['website'].str.extract('(.*?)\.')
I have the following dataframe of securities and computed a 'liquidity score' in the last column, where 1 = liquid, 2 = less liquid, and 3 = illiquid. I want to group the securities (dynamically) by their liquidity. Is there a way to group them and include some kind of header for each group? How can this be best achieved. Below is the code and some example, how it is supposed to look like.
import pandas as pd
df = pd.DataFrame({'ID':['XS123', 'US3312', 'DE405'], 'Currency':['EUR', 'EUR', 'USD'], 'Liquidity score':[2,3,1]})
df = df.sort_values(by=["Liquidity score"])
print(df)
# 1 = liquid, 2 = less liquid,, 3 = illiquid
Add labels for liquidity score
The following replaces labels for numbers in Liquidity score:
df['grp'] = df['Liquidity score'].replace({1:'Liquid', 2:'Less liquid', 3:'Illiquid'})
Headers for each group
As per your comment, find below a solution to do this.
Let's illustrate this with a small data example.
df = pd.DataFrame({'ID':['XS223', 'US934', 'US905', 'XS224', 'XS223'], 'Currency':['EUR', 'USD', 'USD','EUR','EUR',]})
Insert a header on specific rows using np.insert.
df = pd.DataFrame(np.insert(df.values, 0, values=["Liquid", ""], axis=0))
df = pd.DataFrame(np.insert(df.values, 2, values=["Less liquid", ""], axis=0))
df.columns = ['ID', 'Currency']
Using Pandas styler, we can add a background color, change font weight to bold and align the text to the left.
df.style.hide_index().set_properties(subset = pd.IndexSlice[[0,2], :], **{'font-weight' : 'bold', 'background-color' : 'lightblue', 'text-align': 'left'})
You can add a new column like this:
df['group'] = np.select(
[
df['Liquidity score'].eq(1),
df['Liquidity score'].eq(2)
],
[
'Liquid','Less liquid'
],
default='Illiquid'
)
And try setting as index, so you can filter using the index:
df.set_index(['grouping','ID'], inplace=True)
df.loc['Less liquid',:]
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)
I have a dataframe column which contains all the text below in each row
Symbol(id=15351, ticker=VXX, market=US, currency=USD, type=EQUITY,tick_size=0.010000, lot_size=100, contract_size=0, rate=None)
I am trying to extract only after ticker=, which gives VXX
I tried
df['symbolcolumn'] = df['symbolcolumn'].str.split(',market', expand=True)
But it does not extract only the symbol ticker
Looking for df['symbolcolumn'] = VXX
Can you advise me please?
Ok I managed to do it by
df['symbol'] = df['symbol'].astype(str)
df['symbol'] = df['symbol'].str.split(', market', expand=True)
df['symbol'] = df['symbol'].apply(lambda x: x.split("=")[-1])