I have a 20 x 4000 dataframe in python using pandas. Two of these columns are named Year and quarter. I'd like to create a variable called period that makes Year = 2000 and quarter= q2 into 2000q2
Can anyone help with that?
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问题:
回答1:
dataframe["period"] = dataframe["Year"].map(str) + dataframe["quarter"] 回答2:
df = pd.DataFrame({'Year': ['2014', '2015'], 'quarter': ['q1', 'q2']}) df['period'] = df[['Year', 'quarter']].apply(lambda x: ''.join(x), axis=1) Yields this dataframe
Year quarter period 0 2014 q1 2014q1 1 2015 q2 2015q2 This method generalizes to an arbitrary number of string columns by replacing df[['Year', 'quarter']] with any column slice of your dataframe, e.g. df.iloc[:,0:2].apply(lambda x: ''.join(x), axis=1).
You can check more information about apply() method here
回答3:
yet another ways to do this:
df['period'] = df['Year'].astype(str) + df['quarter'] or bit slower:
df['period'] = df[['Year','quarter']].astype(str).sum(axis=1) Let's test it on 200K rows DF:
In [250]: df Out[250]: Year quarter 0 2014 q1 1 2015 q2 In [251]: df = pd.concat([df] * 10**5) In [252]: df.shape Out[252]: (200000, 2) UPDATE: new timings using Pandas 0.19.0
Timing without CPU/GPU optimization (sorted from fastest to slowest):
In [107]: %timeit df['Year'].astype(str) + df['quarter'] 10 loops, best of 3: 131 ms per loop In [106]: %timeit df['Year'].map(str) + df['quarter'] 10 loops, best of 3: 161 ms per loop In [108]: %timeit df.Year.str.cat(df.quarter) 10 loops, best of 3: 189 ms per loop In [109]: %timeit df.loc[:, ['Year','quarter']].astype(str).sum(axis=1) 1 loop, best of 3: 567 ms per loop In [110]: %timeit df[['Year','quarter']].astype(str).sum(axis=1) 1 loop, best of 3: 584 ms per loop In [111]: %timeit df[['Year','quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1) 1 loop, best of 3: 24.7 s per loop Timing using CPU/GPU optimization:
In [113]: %timeit df['Year'].astype(str) + df['quarter'] 10 loops, best of 3: 53.3 ms per loop In [114]: %timeit df['Year'].map(str) + df['quarter'] 10 loops, best of 3: 65.5 ms per loop In [115]: %timeit df.Year.str.cat(df.quarter) 10 loops, best of 3: 79.9 ms per loop In [116]: %timeit df.loc[:, ['Year','quarter']].astype(str).sum(axis=1) 1 loop, best of 3: 230 ms per loop In [117]: %timeit df[['Year','quarter']].astype(str).sum(axis=1) 1 loop, best of 3: 230 ms per loop In [118]: %timeit df[['Year','quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1) 1 loop, best of 3: 9.38 s per loop 回答4:
The method cat() of the .str accessor works really well for this:
>>> import pandas as pd >>> df = pd.DataFrame([["2014", "q1"], ... ["2015", "q3"]], ... columns=('Year', 'Quarter')) >>> print(df) Year Quarter 0 2014 q1 1 2015 q3 >>> df['Period'] = df.Year.str.cat(df.Quarter) >>> print(df) Year Quarter Period 0 2014 q1 2014q1 1 2015 q3 2015q3 cat() even allows you to add a separator so, for example, suppose you only have integers for year and period, you can do this:
>>> import pandas as pd >>> df = pd.DataFrame([[2014, 1], ... [2015, 3]], ... columns=('Year', 'Quarter')) >>> print(df) Year Quarter 0 2014 1 1 2015 3 >>> df['Period'] = df.Year.astype(str).str.cat(df.Quarter.astype(str), sep='q') >>> print(df) Year Quarter Period 0 2014 1 2014q1 1 2015 3 2015q3 回答5:
Use of a lamba function this time with string.format().
import pandas as pd df = pd.DataFrame({'Year': ['2014', '2015'], 'Quarter': ['q1', 'q2']}) print df df['YearQuarter'] = df[['Year','Quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1) print df Quarter Year 0 q1 2014 1 q2 2015 Quarter Year YearQuarter 0 q1 2014 2014q1 1 q2 2015 2015q2 This allows you to work with non-strings and reformat values as needed.
import pandas as pd df = pd.DataFrame({'Year': ['2014', '2015'], 'Quarter': [1, 2]}) print df.dtypes print df df['YearQuarter'] = df[['Year','Quarter']].apply(lambda x : '{}q{}'.format(x[0],x[1]), axis=1) print df Quarter int64 Year object dtype: object Quarter Year 0 1 2014 1 2 2015 Quarter Year YearQuarter 0 1 2014 2014q1 1 2 2015 2015q2 回答6:
As your data are inserted to a dataframe, this command should solve your problem:
df['period'] = df[['Year', 'quarter']].apply(lambda x: ' '.join(x.astype(str)), axis=1) 回答7:
Although the @silvado answer is good if you change df.map(str) to df.astype(str) it will be faster:
import pandas as pd df = pd.DataFrame({'Year': ['2014', '2015'], 'quarter': ['q1', 'q2']}) In [131]: %timeit df["Year"].map(str) 10000 loops, best of 3: 132 us per loop In [132]: %timeit df["Year"].astype(str) 10000 loops, best of 3: 82.2 us per loop 回答8:
Here is an implementation that I find very versatile:
In [1]: import pandas as pd In [2]: df = pd.DataFrame([[0, 'the', 'quick', 'brown'], ...: [1, 'fox', 'jumps', 'over'], ...: [2, 'the', 'lazy', 'dog']], ...: columns=['c0', 'c1', 'c2', 'c3']) In [3]: def str_join(df, sep, *cols): ...: from functools import reduce ...: return reduce(lambda x, y: x.astype(str).str.cat(y.astype(str), sep=sep), ...: [df[col] for col in cols]) ...: In [4]: df['cat'] = str_join(df, '-', 'c0', 'c1', 'c2', 'c3') In [5]: df Out[5]: c0 c1 c2 c3 cat 0 0 the quick brown 0-the-quick-brown 1 1 fox jumps over 1-fox-jumps-over 2 2 the lazy dog 2-the-lazy-dog 回答9:
more efficient is
def concat_df_str1(df): """ run time: 1.3416s """ return pd.Series([''.join(row.astype(str)) for row in df.values], index=df.index) and here is a time test:
import numpy as np import pandas as pd from time import time def concat_df_str1(df): """ run time: 1.3416s """ return pd.Series([''.join(row.astype(str)) for row in df.values], index=df.index) def concat_df_str2(df): """ run time: 5.2758s """ return df.astype(str).sum(axis=1) def concat_df_str3(df): """ run time: 5.0076s """ df = df.astype(str) return df[0] + df[1] + df[2] + df[3] + df[4] + \ df[5] + df[6] + df[7] + df[8] + df[9] def concat_df_str4(df): """ run time: 7.8624s """ return df.astype(str).apply(lambda x: ''.join(x), axis=1) def main(): df = pd.DataFrame(np.zeros(1000000).reshape(100000, 10)) df = df.astype(int) time1 = time() df_en = concat_df_str4(df) print('run time: %.4fs' % (time() - time1)) print(df_en.head(10)) if __name__ == '__main__': main() final, when sum(concat_df_str2) is used, the result is not simply concat, it will trans to integer.
回答10:
As many have mentioned previously, you must convert each column to string and then use the plus operator to combine two string columns. You can get a large performance improvement by using NumPy.
%timeit df['Year'].values.astype(str) + df.quarter 71.1 ms ± 3.76 ms per loop (mean ± std. dev. of 7 runs, 10 loops each) %timeit df['Year'].astype(str) + df['quarter'] 565 ms ± 22.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) 回答11:
Use .combine_first.
df['Period'] = df['Year'].combine_first(df['Quarter']) 回答12:
def madd(x): """Performs element-wise string concatenation with multiple input arrays. Args: x: iterable of np.array. Returns: np.array. """ for i, arr in enumerate(x): if type(arr.item(0)) is not str: x[i] = x[i].astype(str) return reduce(np.core.defchararray.add, x) For example:
data = list(zip([2000]*4, ['q1', 'q2', 'q3', 'q4'])) df = pd.DataFrame(data=data, columns=['Year', 'quarter']) df['period'] = madd([df[col].values for col in ['Year', 'quarter']]) df Year quarter period 0 2000 q1 2000q1 1 2000 q2 2000q2 2 2000 q3 2000q3 3 2000 q4 2000q4