Question

我试图创建一个列'mdiff'，计算'POP ...'列的最大值和最小值之间的差。它给了我一个语法错误，其中：res = df1.nlargest（1）（我试图在这里获取最大值）。我不确定这里有什么问题吗？数据是具有列CTYNAME, POPOPESTIMATE2010,..2011,..2012,..2013,..2014..2015.

的数据框

def maxdiff():
    rows =['POPESTIMATE2010',
    'POPESTIMATE2011',
    'POPESTIMATE2012',
    'POPESTIMATE2013',
    'POPESTIMATE2014',
    'POPESTIMATE2015']
    df1=df.where(df['SUMLEV']==50).set_index(['CTYNAME']).dropna()
    df1['mdiff']=df1.apply(lambda x: abs(np.max(x[rows]-np.(min(x[rows])), 
    axis=1)
    dmax=df1.nlargest(1,'mdiff')   

    return dmax

Answer 1

您在lambda表达式中缺少右括号：

def maxdiff():
    rows = [
        "POPESTIMATE2010",
        "POPESTIMATE2011",
        "POPESTIMATE2012",
        "POPESTIMATE2013",
        "POPESTIMATE2014",
        "POPESTIMATE2015",
    ]
    df1 = (
        df.where(df["SUMLEV"] == 50)
        .set_index(["CTYNAME"])
        .dropna()
    )
    df1["mdiff"] = df1.apply(
        lambda x: abs(np.max(x[rows] - np.min[rows])),  # <-- One more `)` here!
        axis=1
    )
    dmax = df1.nlargest(1, "mdiff")

    return dmax

熊猫数据框中的语法无效

1 个答案: