如何使用python中的现有列以其他列为条件创建新列

时间:2017-09-14 13:46:41

标签: python r python-3.x pandas dataframe

我在 python 中有dataframe,看起来像这样

dt = pd.DataFrame({"language1": ["english", "english123", "ingles", "ingles123", "14.0", "13", "french"],
                   "language2": ["englesh", "english123", "ingles", "ingles123", "14", "13", "french"],
                   "language3": ["englesh", "engl", "ingles", "ingles123", "14", "13", "spanish"]})

我想做的是复制 R 代码,但在 python

dt[,language4:=ifelse(!language1%in%c("french"),paste0(language2,"_win"),paste0(language3,"_lose"))]

我尝试了这个,但它不起作用

dt['language4'] = dt.apply(lambda x: ~x['language1'].isin(['french']), x['language2'] + "_win", x['language3']+"_lose")

所以我想出了这个

dt.loc[~dt['language1'].isin(["french"]),'language4'] = surv_dt_sd['language2'] + \
    "_win"

但我不知道如何在一行中实现else

2 个答案:

答案 0 :(得分:2)

numpy.where可以在这里工作:

dt['language4'] = np.where("french" not in dt['language1'], dt['language2'] + '_win', dt['language2'] + '_lose')

答案 1 :(得分:0)

我也可以分享“apply + ifelse”解决方案:

dt['language4'] = dt.apply(lambda x: x['language2'] + "_win" if not('french' in x['language1']) else x['language3']+"_lose", axis=1)
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