我有一本字典（数组）。我需要他的外表和照片中的一样。怎么做？
我的代码
array = {1: {2016: 22818888925.021294, 2017: 23215791927.55883, 2018: 24241415053.75489, 2019: 25526831987.89783, 2007: 18183059018.027966, 2008: 16868527277.51174, 2009: 18331809529.53519, 2010: 19724006864.797657, 2011: 20392912788.825798, 2012: 19580849153.840134, 2013: 22084393933.539394, 2014: 22030084567.47173, 2015: 21700822555.267574}, 2: {2016: 15081873888.10038, 2017: 15377071873.462284, 2018: 16253879799.12362, 2019: 16517025291.019705, 2007: 10038529774.615318, 2008: 11611749842.042978, 2009: 12040185617.510305, 2010: 12508816879.258118, 2011: 13711111085.100977, 2012: 13219008951.675236, 2013: 14209627046.021126, 2014: 15371996764.61913, 2015: 14417938720.997215}, 6: {2016: 0.6609381349660216, 2017: 0.6623539667069717, 2018: 0.6705004539991144, 2019: 0.6470456380505956, 2007: 0.5520814602571774, 2008: 0.6883677306864346, 2009: 0.6567919876164886, 2010: 0.6341924825418297, 2011: 0.6723468700662476, 2012: 0.6750988605150847, 2013: 0.6434239077958611, 2014: 0.6977729348945159, 2015: 0.6643959547744175}})
print pd.DataFrame.from_dict(mass_factor)
我的代码的输出
1 2 6
2007 1.818306e+10 1.003853e+10 0.552081
2008 1.686853e+10 1.161175e+10 0.688368
2009 1.833181e+10 1.204019e+10 0.656792
2010 1.972401e+10 1.250882e+10 0.634192
2011 2.039291e+10 1.371111e+10 0.672347
2012 1.958085e+10 1.321901e+10 0.675099
2013 2.208439e+10 1.420963e+10 0.643424
2014 2.203008e+10 1.537200e+10 0.697773
2015 2.170082e+10 1.441794e+10 0.664396
2016 2.281889e+10 1.508187e+10 0.660938
2017 2.321579e+10 1.537707e+10 0.662354
2018 2.424142e+10 1.625388e+10 0.670500
2019 2.552683e+10 1.651703e+10 0.647046
如果您使用以下代码，您应该得到您想要的：
new_df = pd.DataFrame(df.unstack()).transpose()
print(new_df)
该作品由你原来的DF与多指标+年通过转换成一连串当前所有列pd.DataFrame.unstack，然后通过喂养串联成pd.DataFrame转换所产生的系列回一个数据帧与一排（导致一长列，标题为 0)，然后使用pd.transpose将该列转置为您想要的一长行。
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